RD Sharma Class 12 Exercise 18.19 Indefinite Integrals Solutions Maths

RD Sharma Class 12 Exercise 18.19 Indefinite Integrals Solutions Maths - Download PDF Free Online

Edited By Kuldeep Maurya | Updated on Jan 24, 2022 12:20 PM IST

RD Sharma Class 12th Exercise 18.19 solutions are the best choice for a student because of the perfect answers to all the questions of RD Sharma which is considered as maths bible. The team always looks forward to finding multiple ways to solve these questions making the journey easy for every student. This particular exercise has 18 questions that are formatted by a team of experts. RD Sharma Solutions Rd Sharma Class 12th Exercise 18.19 discusses completing squaring method to solve quadratic equation, integrating the function, which includes use of fundamental integration formulae to explain standard results on integration.

JEE Main 2025: Sample Papers | Mock Tests | PYQs | Study Plan 100 Days

JEE Main 2025: Maths Formulas | Study Materials

JEE Main 2025: Syllabus | Preparation Guide | High Scoring Topics

This Story also Contains

RD Sharma Class 12 Solutions Chapter18 Indefinite Integrals - Other Exercise
Indefinite Integrals Excercise:18.19
RD Sharma Chapter-wise Solutions

RD Sharma Class 12 Solutions Chapter18 Indefinite Integrals - Other Exercise

Indefinite Integrals Excercise:18.19

Indefinite Integrals Exercise 18.19 Question 1

Answer:

= \frac{1}{2} [\log | x^{2} + 3 x + 2 |] - \frac{3}{2} [\log | \frac{x + 1}{x + 2} |] + c

Hint Using Integration method
Given: $\int \frac{x}{x^{2} + 3 x + 2} d x$
Solution: Let,

x^{2} + 3 x + 2 = t

(2 x + 3) d x = d t

\begin{aligned} I = \frac{2}{2} \int \frac{x}{x^{2} + 3 x + 2} d x \\ = \frac{1}{2} \int \frac{2 x}{x^{2} + 3 x + 2} d x \\ = \frac{1}{2} \int \frac{2 x + 3 - 3}{x^{2} + 3 x + 2} d x \\ = \frac{1}{2} [\int \frac{2 x + 3}{x^{2} + 3 x + 2} d x - \int \frac{3}{x^{2} + 3 x + 2} d x] \end{aligned}

= \frac{1}{2} [\frac{d t}{t} - 3 \int \frac{d x}{(x + 1) (x + 2)}]

[\begin{matrix} ∵ (x^{2} + 3 x + 2) \\ = (x^{2} + 2 x + x + 2) \\ = (x (x + 2) + 1 (x + 2)) \\ = (x + 1) (x + 2) \end{matrix}]

= \frac{1}{2} [\log t - 3 \int (\frac{1}{x + 1} - \frac{1}{x + 2}) d x]

[\int \frac{1}{x} d x = \log x + c]

= \frac{1}{2} [\log (x^{2} + 3 x + 2) - 3 (\log (x + 1) - \log (x + 2)] + c

= \frac{1}{2} [\log (x^{2} + 3 x + 2) - 3 \log (\frac{x + 1}{x + 2})] + c

= \frac{1}{2} [\log (x^{2} + 3 x + 2)] - \frac{3}{2} [\log (\frac{x + 1}{x + 2})] + c

Indefinite Integrals Exercise 18.18 Question 2

Answer: $\frac{1}{2} \log | x^{2} + x + 3 | + \frac{1}{\sqrt{11}} \tan^{- 1} (\frac{2 x + 1}{\sqrt{11}}) + c$
Hint: Integrate by ILATE
Given: $\int \frac{x + 1}{x^{2} + x + 3} d x$
Solution:
$\begin{aligned} I_{1} = \frac{1}{2} \int \frac{2 x + 2}{x^{2} + x + 3} d x \\ I_{1} = \frac{1}{2} \int \frac{(2 x + 1) + 1}{x^{2} + x + 3} d x \end{aligned}$
$= \frac{1}{2} \int \frac{2 x + 1}{x^{2} + x + 3} d x + \frac{1}{2} \int \frac{1}{x^{2} + x + 3} d x$
$\begin{aligned} I_{1} & = \frac{1}{2} \int \frac{2 x + 2}{x^{2} + x + 3} d x \\ I_{1} & = \frac{1}{2} \int \frac{(2 x + 1) + 1}{x^{2} + x + 3} d x \\ = \frac{1}{2} \int \frac{2 x + 1}{x^{2} + x + 3} d x + \frac{1}{2} \int \frac{1}{x^{2} + x + 3} d x \end{aligned}$ $[\int \frac{d t}{t} = \ln t + c,]$
Let,

x^{2} + x + 3 = t

(2 x + 1) d x = d t

I_{1} = \log | x^{2} + x + 3 | + C

I_{2} = \int \frac{1}{x^{2} + x + 3} d x

= \int \frac{1}{{(x + \frac{1}{2})}^{2} - \frac{1}{4} + 3} d x

= \int \frac{1}{{(x + \frac{1}{2})}^{2} + \frac{11}{4}} d x

[\begin{matrix} x + \frac{1}{2} = t \Rightarrow d x = d t \\ \sqrt{\frac{11}{4}} = a = \frac{\sqrt{11}}{2} \end{matrix}]

= \frac{1}{\frac{\sqrt{11}}{2}} \tan^{- 1} (\frac{x + \frac{1}{2}}{\frac{\sqrt{11}}{2}}) + C_{2}

[\int \frac{1}{t^{2} + a^{2}} d t = \frac{1}{a} \tan^{- 1} (\frac{t}{a}) + C_{2}]

I_{2} = \frac{2}{\sqrt{11}} \tan^{- 1} \frac{(2 x + 1)}{\sqrt{11}} + C_{2}

I = \frac{1}{2} (I_{1} + I_{2})

= \frac{1}{2} [\log | x^{2} + x + 3 | + \frac{2}{\sqrt{11}} \tan^{- 1} (\frac{2 x + 1}{\sqrt{11}})] + c

= \frac{1}{2} \log | x^{2} + x + 3 | + \frac{1}{\sqrt{11}} \tan^{- 1} (\frac{2 x + 1}{\sqrt{11}}) + c

Indefinite Integrals Exercise 18.19 Question 3

Answer: $- \frac{2}{\sqrt{5}} \log | \frac{x + 1 - \sqrt{5}}{x + 1 + \sqrt{5}} | + \frac{1}{2} \log | x^{2} + 2 x - 4 | + c$
Hint Find value of A and B
Given: $\int \frac{x - 3}{x^{2} + 2 x - 4} d x$
Solution: Let

x - 3 = A + B \frac{d}{d x} (x^{2} + 2 x - 4)

x - 3 = A + B (2 x + 2)

x - 3 = A + 2 B x + 2 B

On comparing,

[\begin{matrix} x = 2 B x \Rightarrow B = \frac{1}{2} \\ - 3 = A + 2 B \Rightarrow - 3 - 2 \times \frac{1}{2} = A \Rightarrow A = - 4 \end{matrix}]

= - 4 \int \frac{d x}{x^{2} + 2 x - 4} + \frac{1}{2} \int \frac{\frac{d}{d x} (x^{2} + 2 x - 4)}{x^{2} + 2 x - 4}

[\begin{array}{l} x^{2} + 2 x + 1 - 4 - 1 \\ (x + 1)^{2} - (\sqrt{5})^{2} \end{array}]

= - 4 \int \frac{d x}{(x + 1)^{2} - (\sqrt{5})^{2}} + \frac{1}{2} \log | x^{2} + 2 x - 4 |

[\int \frac{d x}{x^{2} - a^{2}} = \frac{1}{2 a} \log | \frac{x - a}{x + a} | + c]

= - \frac{2}{\sqrt{5}} \log | \frac{x + 1 - \sqrt{5}}{x + 1 + \sqrt{5}} | + \frac{1}{2} \log | x^{2} + 2 x - 4 | + c

Indefinite Integrals Exercise 18.19 Question 4

Answer:
$I = \log | x^{2} + 6 x + 13 | - \frac{9}{2} \tan^{- 1} (\frac{x + 3}{2}) + c$
Hint: Using Formula

\int \frac{d x}{x^{2} - a^{2}} = \frac{1}{a} \tan^{- 1} (\frac{x}{a}) + c & \int \frac{d t}{t} = \log t + c

Given: $\int \frac{2 x - 3}{x^{2} + 6 x + 13} d x$
Solution: $\frac{d}{d x} (x^{2} + 6 x + 13) = 2 x + 6$
Let,

2 x - 3 = A (2 x + 6) + B

2 x - 3 = 2 A x + 6 A + B

On comparing,

2 x = 2 A x => A = 1

6 A + B = - 3 => B = - 9

I = \int \frac{2 x + 6}{x^{2} + 6 x + 13} d x - 9 \int \frac{d x}{x^{2} + 6 x + 13}

Let, x^{2} + 6 x + 13 = t

(2 x + 6) d x = d t

I = \int \frac{d t}{t} - 9 \int \frac{d x}{x^{2} + 2 (3) x + 3^{2} - 3^{2} + 13}

I = \int \frac{d t}{t} - 9 \int \frac{d x}{(x + 3)^{2} - 9 + 13}

I = \int \frac{d t}{t} - 9 \int \frac{d x}{(x + 3)^{2} - 4}

I = \int \frac{d t}{t} - 9 \int \frac{d x}{(x + 3)^{2} - 2^{2}}

[\int \frac{d x}{x^{2} - a^{2}} = \frac{1}{a} \tan^{- 1} (\frac{x}{a}) + c & \int \frac{d t}{t} = \log t + c]

I = \log | t | - 9 \times \frac{1}{2} \tan^{- 1} (\frac{x + 3}{2}) + c

I = \log | x^{2} + 6 x + 13 | - \frac{9}{2} \tan^{- 1} (\frac{x + 3}{2}) + c

Indefinite Integrals Exercise 18.19 Question 5

Answer:

- \frac{\sqrt{5}}{15} \tan^{- 1} | \frac{3 x - 2}{\sqrt{5}} | + \frac{1}{2} \log | x^{2} - \frac{4}{3} x + 1 | + c

Hint: Using Formula

\int \frac{d x}{x^{2} - a^{2}} = \frac{1}{a} \tan^{- 1} (\frac{x}{a}) + c & \int \frac{d t}{t} = \log t + c

Given: $\int \frac{x - 1}{3 x^{2} - 4 x + 3} d x$
Solution:
$\int \frac{x - 1}{3 x^{2} - 4 x + 3} d x$
$\frac{1}{9} \int \frac{x - 1}{x^{2} - \frac{4 x}{3} + 1} d x$
$x - 1 = A + B \frac{d}{d x} (x^{2} - \frac{4 x}{3} + 1)$
$x - 1 = A + B (2 x - \frac{4}{3})$ $[\begin{array}{l} x = 2 B x \Rightarrow B = \frac{1}{2} \\ - 1 = A - \frac{4}{3} B \Rightarrow - 1 + \frac{4}{3} \times \frac{1}{2} = A \\ - 1 + \frac{2}{3} = A \Rightarrow A = \frac{- 1}{3} \end{array}]$
$I = \frac{- 1}{3 \times 3} \int \frac{d x}{x^{2} - \frac{4}{3} x + 1} + \frac{1}{3 \times 2} \int \frac{6 x - 4}{x^{2} - \frac{4}{3} x + 1} d x$
$I = \frac{- 1}{9} \int \frac{d x}{x^{2} - \frac{4}{3} x + 1} + \frac{1}{6} \int \frac{d t}{t}$ $[\begin{matrix} Let, x^{2} - \frac{4}{3} x + 1 = t \\ (2 x - \frac{4}{3}) d x = d t \\ \frac{(6 x - 4) d x}{3} = d t \end{matrix}]$
$I = \frac{- 1}{9} \int \frac{d x}{x^{2} - \frac{4}{3} x + \frac{4}{9} + 1 - \frac{4}{9}} + \frac{1}{2} \int \frac{d t}{t}$
$I = - \frac{1}{9} \int \frac{d x}{{(x - \frac{2}{3})}^{2} + {(\frac{\sqrt{5}}{8})}^{2}} + \frac{1}{2} \int \frac{d t}{t}$ $[\int \frac{d x}{x^{2} - a^{2}} = \frac{1}{a} \tan^{- 1} (\frac{x}{a}) + c & \int \frac{d t}{t} = \log t + c]$
$I = \frac{- 1}{9} \times \frac{1}{\frac{\sqrt{5}}{3}} \tan^{- 1} | \frac{x - \frac{2}{3}}{\frac{\sqrt{5}}{3}} | + \frac{1}{2} \log | x^{2} - \frac{4}{3} x + 1 | + c$
$I = \frac{- 1}{3 \sqrt{5}} \tan^{- 1} | \frac{3 x - 2}{\sqrt{5}} | + \frac{1}{2} \log | x^{2} - \frac{4}{3} x + 1 | + c$
$I = \frac{- 3 \sqrt{5}}{3 \sqrt{5} \times 3 \sqrt{5}} \tan^{- 1} | \frac{3 x - 2}{\sqrt{5}} | + \frac{1}{2} \log | x^{2} - \frac{4}{3} x + 1 | + c$
$I = \frac{- \sqrt{5}}{15} \tan^{- 1} | \frac{3 x - 2}{\sqrt{5}} | + \frac{1}{2} \log | x^{2} - \frac{4}{3} x + 1 | + c$

Indefinite Integrals Exercise 18.19 Question 6

Answer: $\frac{- 4}{3} \ln | x - 2 | - \frac{2}{3} \ln | x + 1 | + c$
Hint: You must know about how to solve integration
Given: $\int \frac{2 x}{2 + x - x^{2}} d x$
Solution: $\int \frac{2 x}{2 + x - x^{2}} d x$
$= - \int \frac{2 x}{x^{2} - x - 2} d x$
$= - \int \frac{2 x}{(x - 2) (x + 1)} d x$
$\frac{A}{(x - 2)} + \frac{B}{(x + 1)} = \frac{2 x}{(x - 2) (x + 1)}$
$A (x + 1) + B (x + 2) = 2 x$
$A x + A + B x + 2 B = 2 x$
On comparing,

(\begin{array}{cc} A x + B x = 2 x & A - 2 B = 0 \\ A + B = 2 & A = 2 B \end{array})

2 B + B = 2

B = \frac{2}{3}, A = \frac{4}{3}

\int - \frac{4}{3} (\frac{1}{x - 2}) d x - \frac{2}{3} \int \frac{1}{x + 1} d x

= - \frac{4}{3} \log | x - 2 | - \frac{2}{3} \log | x + 1 | + c

Indefinite Integrals Exercise 18.19 Question 7

Answer:

- \frac{1}{2} \ln (3 x^{2} + 4 x + 2) + \frac{3}{\sqrt{2}} \tan^{- 1} (\frac{3 x + 2}{\sqrt{2}}) + c

Hint:Use

\int \frac{P n + q}{a x^{2} + b x + c}

Given: $\int \frac{1 - 3 x}{3 x^{2} + 4 x + 2} d x$
Solution:
$I = \int \frac{1 - 3 x}{3 x^{2} + 4 x + 2} d n$
$\int \frac{P n + q}{a x^{2} + b x + c} d x$
$Numerator = A (\frac{d}{d x} (deno min ator)) + B$
$1 - 3 x = A (\frac{d}{d x} (3 x^{2} + 4 x + 2)) + B$
$1 - 3 x = A (6 x + 4) + B$
$1 - 3 x = 6 A x + 4 A + B$
On comparing,

6 A = - 3 \Rightarrow A = - \frac{3}{6} \Rightarrow A = \frac{- 1}{2}

4 A + B = 1 \Rightarrow 4 (\frac{- 1}{2}) + B = 1 \Rightarrow - 2 + B = 1 \Rightarrow B = 3

I = \int \frac{(- \frac{1}{2}) (6 x + 4) + 3}{3 x^{2} + 4 x + 2} d x

I = \int - \frac{1}{2} \frac{(6 x + 4)}{3 x^{2} + 4 x + 2} d x + \int \frac{3}{3 x^{2} + 4 x + 2} d x

= \frac{- 1}{2} \int \frac{(6 x + 4)}{3 x^{2} + 4 x + 2} d x + \int \frac{3}{3 (x^{2} + \frac{4}{3} x + \frac{2}{3})} d x

= \frac{- 1}{2} \int \frac{(6 x + 4)}{3 x^{2} + 4 x + 2} d x + \int \frac{1}{x^{2} + \frac{4}{3} x + \frac{4}{9} - \frac{4}{9} + \frac{2}{3}} d x

= \frac{- 1}{2} \int \frac{\frac{d}{d x} (3 x^{2} + 4 x + 2)}{3 x^{2} + 4 x + 2} d x + \int \frac{1}{{(x + \frac{2}{3})}^{2} + {(\frac{\sqrt{2}}{3})}^{2}} d x

\begin{aligned} 3 x^{2} + 4 x + 2 = t \\ \frac{d (3 x^{2} + 4 x + 2)}{d x} = d t \end{aligned}

= - \frac{1}{2} \int \frac{d t}{t} + \frac{1}{\frac{\sqrt{2}}{3}} [\frac{\tan^{- 1} x + \frac{2}{3}}{\frac{\sqrt{2}}{3}}] [\int \frac{1}{x^{2} + a^{2}} = \frac{1}{a} \tan^{- 1} \frac{x}{2} + c]

= - \frac{1}{2} \ln (3 x^{2} + 4 x + 2) + \frac{3}{\sqrt{2}} \tan^{- 1} (\frac{3 x + 2}{\sqrt{2}}) + c

Indefinite Integrals Exercise 18.19 Question 8

Answer: $\log | x^{2} - x - 2 | + 2 \log | \frac{x - 2}{x + 1} | + c$
Hint: You must know about how to solve integration
Given: $\int \frac{2 x + 5}{x^{2} - x - 2} d x$
Solution:
Let

I = \int \frac{2 x + 5}{x^{2} - x - 2} d x

= \int \frac{2 x - 1 + 6}{x^{2} - x - 2} d x

= \int \frac{2 x - 1}{x^{2} - x - 2} d x + \int \frac{6}{x^{2} - x - 2} d x

I_{1} = \int \frac{2 x - 1}{x^{2} - x - 2} d x

Let

x^{2} - x - 2 = t

\Rightarrow (2 x - 1) d x = d t

\begin{aligned} I_{1} = \int \frac{d t}{t} = \log | t | + c_{1} \\ = \log | x^{2} - x - 2 | + c_{1} \end{aligned}

\begin{aligned} I_{2} = \int \frac{6}{x^{2} - x - 2} d x \\ = 6 \int \frac{d x}{x^{2} - x + \frac{1}{4} - \frac{1}{4} - 2} \\ = 6 \int \frac{d x}{{(x - \frac{1}{2})}^{2} - {(\frac{3}{2})}^{2}} [\int \frac{d x}{x^{2} - a^{2}} = \frac{1}{2 a} \log | \frac{x - a}{x + a} | + c] \\ = 6 \times \frac{1}{3} \log | \frac{x - \frac{1}{2} - \frac{3}{2}}{x - \frac{1}{2} + \frac{3}{2}} | + c_{2} \\ = 2 \log | \frac{2 x - 4}{2 x + 2} | + c_{2} \end{aligned}

So,

I = I_{1} + I_{2}

= \log | x^{2} - x - 2 | + 2 \log | \frac{x - 2}{x + 1} | + c

Indefinite Integrals Exercise 18.19 Question 9

Answer:

\frac{a}{4} \ln | x^{4} + c^{2} | + \frac{b}{2 c} \tan^{- 1} \frac{x^{2}}{c} + f

Hint : Find

I_{1} a n d I_{2}

Given: $\int \frac{a x^{3} + b x}{x^{4} + c^{2}} d x$
Solution: $\int \frac{a x^{3} + b x}{x^{4} + c^{2}} d x$
$\begin{aligned} I & = \int \frac{a x^{3}}{x^{4} + c^{2}} d x + \int \frac{b x}{x^{4} + c^{2}} d x \\ I_{1} & = \int \frac{a x^{3}}{x^{4} + c^{2}} d x \end{aligned}$
Let,

x^{4} + c^{2} = t

\begin{aligned} 4 x^{3} d x = d t \\ x^{3} d x = \frac{1}{4} d t \end{aligned}

\Rightarrow I_{1} = \frac{a}{4} \int \frac{d t}{t} = \frac{a}{4} \ln t + C_{1}

\begin{aligned} I_{2} = \int \frac{b x}{x^{4} + c^{2}} d x \\ Let x^{2} = v \\ 2 x d x = d v \end{aligned}

I_{2} = \frac{b}{2} \int \frac{d v}{v^{2} + c^{2}}

= \frac{b}{2} \tan^{- 1} (\frac{v}{c}) \times \frac{1}{c} + c_{2} [\int \frac{1}{x^{2} + a^{2}} = \frac{1}{a} \tan^{- 1} \frac{x}{2} + c]

I_{2} = \frac{b}{2 c} \tan^{- 1} (\frac{x^{2}}{c}) + c_{2}

\begin{aligned} I = I_{1} + I_{2} \\ I = \frac{a}{4} \ln | x^{4} + c^{2} | + c_{1} + \frac{b}{2 c} \tan^{- 1} (\frac{x^{2}}{c}) + c_{2} \\ c_{1} + c_{2} = f \\ I = \frac{a}{4} \ln | x^{4} + c^{2} | + \frac{b}{2 c} \tan^{- 1} (\frac{x^{2}}{c}) + f \end{aligned}

Indefinite Integrals Exercise 18.19 Question 10

Answer: $\frac{4}{2 - \sin x} + 3 \log | 2 - \sin x | + c$
Hint: Use integration method
Given: $\int \frac{(3 \sin x - 2) \cos x}{5 - \cos^{2} x - 4 \sin x} d x$
Solution: $I = \int \frac{(3 \sin x - 2) \cos x}{5 - \cos^{2} x - 4 \sin x} d x$
$= \int \frac{(3 \sin x - 2) \cos x}{5 - (1 - \sin^{2} x) - 4 \sin x} d x$
$= \int \frac{(3 \sin x - 2) \cos x}{4 + \sin^{2} x - 4 \sin x} d x$
$= \int \frac{(3 \sin x - 2) \cos x}{(2 - \sin x)^{2}} d x$
$\begin{aligned} Let t = 2 - \sin x \\ \sin x = 2 - t \\ d t = - \cos x d x \end{aligned}$
$\begin{aligned} I = \int \frac{3 (2 - t) - 2}{t^{2}} (- d t) \\ = - \int \frac{4 - 3 t}{t^{2}} d t \end{aligned}$
$\begin{aligned} = - 4 \int \frac{d t}{t^{2}} + 3 \int \frac{1}{t} d t \\ I = - 4 (\frac{- 1}{t}) + 3 \log t + c_{1} \\ I = \frac{4}{2 - \sin x} + 3 \log | 2 - \sin x | + c \end{aligned}$

Indefinite Integrals Exercise 18 .19 Question 11

Answer:

\frac{1}{4} \ln (2 x^{2} + 6 x + 5) + \frac{1}{2} \tan^{- 1} (2 x + 3) + c

Hint: Multiply and Divide by 4
Given: $\int \frac{x + 2}{2 x^{2} + 6 x + 5} d x$
Solution:Multiply and divide by 4

\frac{1}{4} \int \frac{4 x + 6}{2 x^{2} + 6 x + 5} d x + \frac{2}{4} \int \frac{d x}{2 x^{2} + 6 x + 5}

.........(1)

\begin{aligned} I_{1} = \int \frac{4 x + 6}{2 x^{2} + 6 x + 5} d x \\ 2 x^{2} + 6 x + 5 = t \\ (4 x + 6) d x = d t \\ I_{1} = \int \frac{d t}{t} = \ln t + c = \ln (2 x^{2} + 6 x + 5) + c_{1} \end{aligned}

\begin{aligned} I_{2} = \frac{1}{2} \int \frac{d x}{2 x^{2} + 6 x + 5} \\ I_{2} = \frac{1}{2} \int \frac{d x}{x^{2} + 3 x + \frac{9}{4} + \frac{5}{2} - \frac{9}{4}} \end{aligned}

\frac{1}{2} \int \frac{d x}{{(x + \frac{3}{2})}^{2} + \frac{1}{4}}

\begin{aligned} (x + \frac{3}{2}) = u \Rightarrow 1 d x = d u \\ I_{2} = \frac{1}{2} \int \frac{d u}{u^{2} + {(\frac{1}{2})}^{2}} \\ \frac{1}{2} [\frac{1}{\frac{1}{2}} \tan^{- 1} (\frac{u}{\frac{1}{2}})] + c_{2} [\int \frac{1}{x^{2} + a^{2}} = \frac{1}{a} \tan^{- 1} \frac{x}{2} \\ \begin{matrix} \tan^{- 1} [2 \tan^{- 1} 2 (x + \frac{3}{2})] + c_{2} \\ I_{2} = \tan^{- 1} (2 x + 3) + c_{2} \\ I = \frac{1}{4} \ln (2 x^{2} + 6 x + 5) + c_{1} + \frac{1}{2} \tan^{- 1} (2 x + 3) + c_{2} \\ I = \frac{1}{4} \ln (2 x^{2} + 6 x + 5) + \frac{1}{2} \tan^{- 1} (2 x + 3) + c \end{matrix} \end{aligned}

Indefinite Integrals Exercise 18.19 Question 12

Answer: $\frac{5}{6} \ln | 3 x^{2} + 2 x + 1 | - \frac{11}{3 \sqrt{2}} \tan^{- 1} \frac{3 x + 1}{\sqrt{2}} + c$
Hint: Find value of M and N
Given: $\int \frac{5 x - 2}{1 + 2 x + 3 x^{2}} d x$
Solution: $I = \int \frac{5 x - 2}{1 + 2 x + 3 x^{2}} d x$
$\begin{aligned} 5 x - 2 = M \frac{d}{d x} (1 + 2 x + 3 x^{2}) + N \\ 5 x - 2 = M (2 + 6 x) + N \\ 5 x - 2 = 2 M + 6 M x + N \end{aligned}$
On comparing,

\begin{aligned} 6 M = 5 \Rightarrow M = \frac{5}{6} \\ 2 M + N = - 2 \Rightarrow 2 (\frac{5}{6}) + N = - 2 \\ \Rightarrow \frac{5}{3} + N = - 2 \Rightarrow N = - 2 - \frac{5}{3} = - \frac{11}{3} \end{aligned}

I = \frac{5}{6} \int \frac{2 + 6 x}{1 + 2 x + 3 x^{2}} d x - \frac{11}{3} \int \frac{d x}{1 + 2 x + 3 x^{2}}

................................(1)

\begin{aligned} I_{1} = \int \frac{2 + 6 x}{1 + 2 x + 3 x^{2}} d x \\ let, t = 1 + 2 x + 3 x^{2} \\ \Rightarrow 2 + 6 x = \frac{d t}{d x} \\ \Rightarrow (2 + 6 x) d x = d t \end{aligned}

d x = \frac{d t}{2 + 6 x}

\begin{aligned} I_{1} = \int \frac{d t}{t} = \log | t | + c \\ I_{2} = \int \frac{d x}{1 + 2 x + 3 x^{2}} \end{aligned}

\begin{aligned} = 3 (x^{2} + 2 \cdot x \cdot \frac{1}{3} + {(\frac{1}{3})}^{2} - {(\frac{1}{3})}^{2} + \frac{1}{3}) \\ = 3 ({(x + \frac{1}{3})}^{2} + \frac{1}{3} - \frac{1}{9}) \\ = 3 ({(x + \frac{1}{3})}^{2} + \frac{3 - 1}{9}) \\ = 3 ({(x + \frac{1}{3})}^{2} + {(\frac{\sqrt{2}}{3})}^{2}) \end{aligned}

From eq 1

I = \frac{5}{6} \int \frac{d t}{t} - \frac{11}{3} \int \frac{d x}{3 {(x + \frac{1}{3})}^{2} + {(\frac{\sqrt{2}}{3})}^{2}}

= \frac{5}{6} \log | t | - \frac{11}{9} \int \frac{d x}{{(x + \frac{1}{3})}^{2} + {(\frac{\sqrt{2}}{3})}^{2}}

\int \frac{d x}{x^{2} + a^{2}} = \frac{1}{a} \tan^{- 1} \frac{x}{a} + c

= \frac{5}{6} \log | 3 x^{2} + 2 x + 1 | - \frac{11}{9} \times \frac{1}{\frac{\sqrt{2}}{3}} \tan^{- 1} \frac{(x + \frac{1}{3})}{\frac{\sqrt{2}}{3}}

I = \frac{5}{6} \log | 3 x^{2} + 2 x + 1 | - \frac{11}{2 \sqrt{3}} \tan^{- 1} \frac{(3 x + 1)}{\sqrt{2}} + c

Indefinite Integrals Exercise 18.19 Question 13

Answer:
$\frac{1}{3} \log | 3 x - 2 | + c$
Hint: Solve integration by equation
Given:

\begin{aligned} \int \frac{x + 5}{3 x^{2} + 13 x - 10} d x \\ = \int \frac{x + 5}{3 x^{2} + 15 x - 2 x - 10} d x \\ = \int \frac{x + 5}{3 x (x + 5) - 2 (x + 5)} d x \end{aligned}

$\begin{aligned} = \int \frac{x + 5}{(3 x - 2) (x + 5)} d x \\ = \int \frac{d x}{3 x - 2} \end{aligned}$ .........(1)
$\begin{aligned} Let, 3 x - 2 = t \\ \begin{array}{r} 3 d x = d t \\ d x = \frac{d t}{3} \end{array} \end{aligned}$
$\begin{aligned} = \int \frac{d t}{3 t} + c \\ = \frac{1}{3} \log | 3 x - 2 | + c \end{aligned}$

Indefinite Integrals Exercise 18.19 Question 14

Answer: $10 \ln | \sin x - 4 | - 7 \ln | \sin x - 3 | + c$
Hint : Let

\sin x = t

Given: $\int \frac{(3 \sin x - 2) \cos x}{13 - \cos^{2} x - 7 \sin x} d x$
Solution: $Let \sin x = t$
$\cos x d x = d t$
$\int \frac{(3 \sin x - 2) \cos x}{13 - (1 - \sin^{2} x) - 7 \sin x} d x$ $(\cos^{2} x = 1 - \sin^{2} x)$
$\int \frac{(3 t - 2) d t}{13 - 1 + t^{2} - 7 t} = \int \frac{(3 t - 2) d t}{t^{2} - 7 t + 12}$ $[\begin{array}{l} t^{2} - 7 t + 12 = (t - 4) (t - 3) \\ a x^{2} + b x + c, x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} \end{array}]$
$\int \frac{3 t - 2}{(t - 4) (t - 3)} d t = \frac{A}{t - 4} + \frac{B}{t - 3}$ ..........(1)
$\begin{aligned} 3 t - 2 = \frac{A}{t - 4} + \frac{B}{t - 3} \\ 3 t - 2 = A (t - 3) + B (t - 4) \\ 3 t - 2 = A t - 3 A + B t - 4 B \end{aligned}$
On comparing,

3 t = At + Bt \Rightarrow A + B = 3 \Rightarrow> A = 3 - B

- 2 = - 3 A - 4 B = 23 A + 4 B = 2 => 3 (3 - B) + 4 B = 2 => B = - 7

On solving both

A = 3 - (- 7) \Rightarrow A = 10

In eqn (1)

\begin{aligned} \int [\frac{10}{t - 4} + \frac{- 7}{t - 3}] d t \\ = & \int \frac{10}{t - 4} d t - \int \frac{7}{t - 3} d t \end{aligned}

\int \frac{1}{t} d t = \ln | t | + c

\begin{aligned} 10 \ln | t - 4 | - 7 l n | t - 3 | + c \\ = 10 \ln | t - 4 | - 7 \ln | t - 3 | + c \\ = 10 \ln | \sin x - 4 | - 7 \ln | \sin x - 3 | + c \end{aligned}

Indefinite Integrals Exercise 18.19 Question 15

Answer:

\log \frac{| 3 x + 4 |}{3} + c

Hint: Solve Integration
Given: $\int \frac{x + 7}{3 x^{2} + 25 x + 28} d x$
Solution:
$\begin{aligned} \int \frac{x + 7}{3 x^{2} + 25 x + 28} d x \\ = \int \frac{x + 7}{3 x^{2} + 21 x + 4 x + 28} d x \\ = \int \frac{x + 7}{3 x (x + 7) + 4 (x + 7)} d x \\ = \int \frac{x + 7}{(3 x + 4) (x + 7)} d x \\ = \int \frac{d x}{3 x + 4} \end{aligned}$
Let,

\begin{aligned} 3 x + 4 = t \\ 3 d x = d t \end{aligned}

d x = \frac{d t}{3}

\begin{aligned} = \int \frac{d t}{3 t} + c \\ = \frac{1}{3} \log | 3 x + 4 | + c \end{aligned}

Indefinite Integrals Exercise 18.19 Question 16

Answer: $\frac{14}{9} \log | x - 3 | + \frac{13}{9} \ln | x + 6 | + c$
Hint: You know about how to solve integration
Given: $\int \frac{3 x + 5}{x^{2} + 3 x - 18} d x$
Explanation:
We have,

\int \frac{3 x + 5}{x^{2} + 3 x - 18} d x

\begin{aligned} \Rightarrow \int \frac{3 x + 5}{x^{2} + 6 x - 3 x - 18} d x \\ \Rightarrow \int \frac{3 x + 5}{x (x + 6) - 3 (x + 6)} d x \\ \Rightarrow \int \frac{3 x + 5}{(x - 3) (x + 6)} d x \end{aligned}

Now, Using Partial fraction,

\begin{aligned} \frac{3 x + 5}{(x - 3) (x + 6)} = \frac{A}{(x + 6)} + \frac{B}{(x - 3)} \\ \frac{3 x + 5}{(x - 3) (x + 6)} = \frac{A (x - 3) + B (x + 6)}{(x - 3) (x + 6)} \\ \frac{3 x + 5}{(x - 3) (x + 6)} = \frac{A x - 3 A + B x + 6 B}{(x - 3) (x + 6)} \\ \Rightarrow \frac{3 x + 5}{(x - 3) (x + 6)} = \frac{(A + B) x - 3 A + 6 B}{(x - 3) (x + 6)} \end{aligned}

On comparing, we got

A + B = 3

......(1)
and

\begin{aligned} - 3 A + 6 B = 5 \\ (A + B = 3) \times - 3 \end{aligned}

........(2)
We get,

- 3 A - 3 B = - 9 \dots . (3)

Subtract (2) from (3)

\begin{aligned} (- 3 A - 3 B) - (- 3 A + 6 B) = (- 9) - (5) \\ - 3 A - 3 B + 3 A - 6 B = - 9 - 5 \\ - 9 B = - 14 \\ B = \frac{14}{9} \end{aligned}

Put value of B in (1)

\begin{aligned} A + B = 3 \\ A + \frac{14}{9} = 3 \\ A = 3 - \frac{14}{9} \Rightarrow \frac{27 - 14}{9} \\ A = \frac{13}{9} & B = \frac{14}{9} \end{aligned}

Substituting value of A and B in

\begin{aligned} \frac{3 x + 5}{(x - 3) (x + 6)} = \frac{A}{(x + 6)} + \frac{B}{(x - 3)} \\ \frac{3 x + 5}{(x - 3) (x + 6)} = \frac{13}{9} \frac{1}{(x + 6)} + \frac{14}{9} \frac{B}{(x - 3)} \end{aligned}

Integrating both sides

\int \frac{3 x + 5}{(x - 3) (x + 6)} d x = \frac{13}{9} \int \frac{1}{(x + 6)} d x + \frac{14}{9} \int \frac{1}{(x - 3)} d x

[\int \frac{1}{x + a} d x = \log | x + a |]

\int \frac{3 x + 5}{(x - 3) (x + 6)} d x = \frac{13}{9} \log | x + 6 | + \frac{14}{9} \log | x - 3 | + c

Indefinite Integrals Exercise 18 .19 Question 17

Answer:

\frac{1}{4} \ln | x^{4} + x^{2} + 1 | - \frac{1}{2 \sqrt{3}} \tan^{- 1} (\frac{2 x^{2} + 1}{\sqrt{3}}) + c

Hint: You know about the integration of

\int x^{3}

Given: $\int \frac{x^{3}}{x^{4} + x^{2} + 1} d x$
Solution: $\int \frac{x^{3}}{x^{4} + x^{2} + 1} d x$
$\begin{aligned} \int \frac{x^{3}}{x^{4} + x^{2} + 1} d x \\ = \frac{1}{2} \int \frac{2 x^{3}}{x^{4} + x^{2} + 1} d x \\ = \frac{1}{2} \int \frac{x^{2}}{x^{4} + x^{2} + 1} 2 x d x \end{aligned}$
Let

x^{2} = t

2 x d x = d t

\begin{aligned} \frac{1}{2} \int \frac{t}{t^{2} + t + 1} d t \\ \Rightarrow \frac{1}{2} \int \frac{2 t + 1 - 1}{2 (t^{2} + t + 1)} d t \\ \Rightarrow \frac{1}{4} \int \frac{2 t + 1}{t^{2} + t + 1} d t - \int \frac{d t}{t^{2} + t + 1} \\ \Rightarrow \frac{1}{4} \int \frac{2 t + 1}{t^{2} + t + 1} d t - \int \frac{d t}{t^{2} + t + \frac{1}{4} - \frac{1}{4}} \end{aligned}

\Rightarrow \frac{1}{4} \int \frac{2 t + 1}{t^{2} + t + 1} d t - \int \frac{d t}{(\frac{3}{4}) + {(t + \frac{1}{2})}^{2}}

Let,

\begin{aligned} t^{2} + 2 t + 1 & = u \\ 2 t + 1 d t & = d u \end{aligned}

\begin{aligned} = \frac{1}{4} \int \frac{d u}{u} - \int \frac{d t}{{(\frac{\sqrt{3}}{2})}^{2} + {(t + \frac{1}{2})}^{2}} \\ = \frac{1}{4} [\ln u - \frac{1}{\frac{\sqrt{3}}{2}} \tan^{- 1} (\frac{t + \frac{1}{2}}{\frac{\sqrt{3}}{2}})] + c \end{aligned}

\begin{aligned} = \frac{1}{4} (\ln (t^{2} + t + 1) - \frac{2}{\sqrt{3}} \tan^{- 1} (\frac{2 t + 1}{\sqrt{3}})) + c \\ = \frac{1}{4} \ln | x^{4} + x^{2} + 1 | - \frac{1}{4} \frac{2}{\sqrt{3}} \tan^{- 1} (\frac{2 x^{2} + 1}{\sqrt{3}}) + c \\ = \frac{1}{4} \ln | x^{4} + x^{2} + 1 | - \frac{1}{2 \sqrt{3}} \tan^{- 1} (\frac{2 x^{2} + 1}{\sqrt{3}}) + c \end{aligned}

Indefinite Integrals Exercise 18.19 Question 18

Answer:

\frac{- \sqrt{5}}{5} \log | \frac{x^{2} + 1 - \sqrt{5}}{x^{2} + 1 + \sqrt{5}} | + \frac{1}{4} \log | x^{4} - 2 x^{2} - 4 | + c

Hint : Let

x^{2} = t

Given: $\int \frac{x^{3} - 3 x}{x^{4} + 2 x^{2} - 4} d x$
Solution: $\int \frac{x (x^{2} - 3)}{{(x^{2})}^{2} + 2 x^{2} - 4} d x$
Let

x^{2} = t

x d x = \frac{d t}{2}

\frac{1}{2} \int \frac{(t - 3)}{t^{2} + 2 t - 4} d t

Multiply and divide by 2

\begin{aligned} = \frac{1}{4} \int \frac{2 t - 6}{t^{2} + 2 t - 4} d t \\ = \frac{1}{4} \int \frac{2 t + 2 - 8}{t^{2} + 2 t - 4} d t \\ = \frac{1}{4} \int \frac{2 t + 2}{t^{2} + 2 t - 4} d t - \frac{1}{4} \int \frac{8}{t^{2} + 2 t - 4} d t \\ = \frac{1}{4} \int \frac{2 t + 2}{t^{2} + 2 t - 4} d t - 2 \int \frac{1}{t^{2} + 2 t - 4} d t \\ = \frac{1}{4} \int \frac{2 t + 2}{t^{2} + 2 t - 4} d t - 2 \int \frac{1}{t^{2} + 2 t + 1 - 5} d t \\ = \frac{1}{4} \int \frac{2 t + 2}{t^{2} + 2 t - 4} d t - 2 \int \frac{1}{(t + 1)^{2} - {\sqrt{5}}^{2}} d t \end{aligned}

I = \frac{1}{4} \log | t^{2} + 2 t - 4 | - 2 \times \frac{1}{2 \sqrt{5}} \log (\frac{t + 1 - \sqrt{5}}{t + 1 + \sqrt{5}}) + c

I = \frac{1}{4} \log | x^{4} + 2 x^{2} - 4 | - \frac{1}{\sqrt{5}} \log (\frac{x^{2} + 1 - \sqrt{5}}{x^{2} + 1 + \sqrt{5}}) + c

Rd Sharma Class 12. Exercise 18.19 The solutions will help each student and they will be able to clarify the concepts and study from the point of view of an exam. The advantages of these solutions are:

Created by experts

Class 12th RD Sharma Chapter 18 Exercise 18.19 Solutions are created by a team of experts making the solutions a reliable source to learn the material. These solutions will always help students to go for the exams confidently and perform brilliantly.

Best source for preparation

JEE Main Highest Scoring Chapters & Topics

Just Study 40% Syllabus and Score upto 100%

Download E-book

Rd Sharma Class 12th Exercise 18.19 solutions being the reliable ones are the best source for preparation. A student can trust these solutions and go for them without worry. The expert team who writes the answers analyses every question to solve it in advanced ways so as to make it easy for students to understand.

NCERT Questions Answered

The questions appearing in board exams are often from NCERT maths textbooks. Teachers also like to set question papers from NCERT books, thus Rd Sharma Class 12th Chapter 18 Exercise 18.19 solutions helps students to prepare for school exams along with boards. The ultimate solution for all maths doubts is the RD Sharma Class 12 Solutions Chapter 18 ex 18.19 Solutions book.

Different approaches to resolve a question

RD Sharma Class 12 Solutions Chapter 18 ex 18.19 allows students to locate opportunity approaches to resolve a question. The principles are connected and solved through specialists in any such manner that it is straightforward for a pupil to apprehend and resolve questions in different approaches.

Great performance in exams:

Solutions assist students do properly in exams. RD Sharma Class 12 Solutions Ex. 18.19 covers all of the critical questions typically requested on exams, assisting the pupil to set up a benchmark rating on exams.

Free of price

Students will locate these types of answers freed from price at Career360, the fine web website online to go to for all of the blessings associated with questions in board checks in addition to aggressive checks. At Career360, a pupil will understand, study and carry out fairly checks. Thousands of college students have benefitted from those answers making Career360 primary desire for each pupil

RD Sharma Chapter-wise Solutions

Frequently Asked Questions (FAQs)

1. How to ace the chapter on indefinite integrals?

The indefinite integral Chapter is difficult and tedious. However, using RD Sharma solutions will help you master the concepts through regular exercise.

2. Why should I study RD Sharma solutions?

RD Sharma solutions will help students score great marks in exams by improving their understanding and learning of mathematics. Every question is created by a team of experts who are knowledgeable and mindful about helping students.

3. Are RD Sharma solutions expensive to own?

No, RD Sharma solutions are actually free of cost and can be owned by anyone. The free copy of the book is always available at Career360.

4. Define Indefinite Integrals in mathematics?

An indefinite integral refers to any integral without an upper and lower limit.

5. What are the different types of Integrals?

The integrals are two types and these are:

Definite Integral
Indefinite Integral

Articles

NIOS Exam Date 2025 for April Exams (Out): Class 10, 12 Theory Date Sheet PDF

Mar 20, 2025

NIOS Hall Ticket 2025 for Class 10th and 12th for April Session, Check Admit Card Here

Mar 19, 2025

NCERT Exemplar Class 10 Maths Solutions Chapter 7 Coordinate Geometry

Apr 02, 2025

NIOS Class 12 Date Sheet 2025 Out (April Session), Check Theory and Practical Exam Time Table

Mar 20, 2025

NIOS 12th Hall Ticket 2025 (April Exam) - Download NIOS Class 12 Admit Card

Mar 20, 2025

NIOS 10th Hall Ticket 2025 April session, Download NIOS 10th Public Exams Admit Card Here

Apr 02, 2025

CBSE Class 11 Economics Syllabus 2025-26 Out: Download Latest Curriculum PDF

Apr 02, 2025

NCERT Exemplar Class 12 Biology Solutions Chapter 16 Environmental Issues

Application Date:24 March,2025 - 23 April,2025

Exam Pattern

Admit Card

Result

View All School Exams

Exams

Top Schools

Products & Resources

NCERT Study Material

Exams

Colleges

Predictors

Resources

Exams

Colleges & Courses

Predictors

Resources

Exams

Colleges

Predictors

Resources

Exams

Colleges

Predictors & E-Books

Resources

Exams

Predictors & Articles

Colleges

Resources

Exams

Colleges

Resources

Exams

Resources

Top Courses & Careers

Colleges

Exams

Colleges

Resources

Quick Links

Exams

Colleges

Resources

Exams

Colleges

Resources

Diploma Colleges

Exams

Resources

Upcoming Events

Other Exams

Exams

Colleges

Top Countries

Student Visas

Exams

Colleges

Upcoming Events

Resources

Engineering Preparation

Medical Preparation

Online Courses

Products

Top Streams

Specializations

Resources

Top Providers

RD Sharma Class 12 Exercise 18.19 Indefinite Integrals Solutions Maths - Download PDF Free Online

RD Sharma Class 12 Solutions Chapter18 Indefinite Integrals - Other Exercise

Indefinite Integrals Excercise:18.19

RD Sharma Chapter-wise Solutions

Frequently Asked Questions (FAQs)

Articles

Upcoming School Exams

Central Board of Secondary Education Class 12th Examination

Andhra Pradesh Residential Junior Colleges Common Entrance Test

Telangana State Secondary School Certificate Examination

Karnataka Secondary School Leaving Certificate Examination

Telangana Residential Junior Colleges Common Entrance Test

Also Read

Study Resources, Applications and Opportunities

NEET/JEE Coaching Scholarship

NEET Most scoring concepts

NEET Previous 10 Year Questions

JEE Main Important Physics formulas