RD Sharma Class 12 Exercise 18.21 Indefinite Integrals Solutions Maths

RD Sharma Class 12 Exercise 18.21 Indefinite Integrals Solutions Maths - Download PDF Free Online

Edited By Kuldeep Maurya | Updated on Jan 24, 2022 12:25 PM IST

Rd Sharma Class 12th Exercise 18.21 is very important because the RD Sharma Book has always been the best source for preparation. Every student studies from RD Sharma books and gets success. Mathematics being a complicated subject, needs best solutions for it which explain every bit of the concept to help students grasp it fast. Rd Sharma Solutions Rd Sharma Class 12th Exercise 18.21 has solved every problem of a student regarding differentiation and integration. This exercise has 18 questions based on quadratic equations, integrating the function, and some important formulae used for evaluating the integrals.

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RD Sharma Class 12 Solutions Chapter18 Indefinite Integrals - Other Exercise

Indefinite Integrals Excercise:18.21

Indefinite Integrals Exercise 18.21 Question 1

Answer:

\sqrt{x^{2} + 6 x + 10} - 3 [\log | (x + 3) + \sqrt{x^{2} + 6 x + 10} |] + c

Given:

\int \frac{x}{\sqrt{x^{2} + 6 x + 10}} d x

Hint: Simplify it and integrate it with the help of formula.
Solution: Let

I = \int \frac{x}{\sqrt{x^{2} + 6 x + 10}} d x

= \int \frac{x + 3 - 3}{\sqrt{x^{2} + 6 x + 10}} d x

......................Adding +3 and -3 in numerator

I = \frac{1}{2} \int \frac{2 (x + 3)}{\sqrt{x^{2} + 6 x + 10}} d x - 3 \int \frac{d x}{\sqrt{x^{2} + 6 x + 10}}

I = I_{1} - I_{2}

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

\begin{aligned} I_{1} = \frac{1}{2} \int \frac{2 (x + 3)}{\sqrt{x^{2} + 6 x + 10}} d x \\ I_{2} = 3 \int \frac{d x}{\sqrt{x^{2} + 6 x + 10}} \end{aligned}

First solve

I_{1}

Let

\begin{aligned} x^{2} + 6 x + 10 = y \\ (2 x + 6) d x = d y \end{aligned}

.........(2)

I_{1} = \frac{1}{2} \int \frac{d y}{\sqrt{y}} = \frac{1}{2} \int (y)^{- \frac{1}{2}} d y

I_{1} = \frac{1}{2} [\frac{y^{\frac{1}{2}}}{\frac{1}{2}}] + c

I_{1} = \sqrt{y + c}

From equation (2)

I_{1} = \sqrt{x^{2} + 6 x + 10} + c

Now

I_{2} = 3 \int \frac{d x}{\sqrt{x^{2} + 6 x + 10}}

x^{2} + 6 x + 10 = x^{2} + 6 x + 9 - 9 + 10

= {(x + 3)}^{2} + 1

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

I_{2} = 3 \log | x + 3 + \sqrt{x^{2} + 6 x + 10} | + c

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

Now, putting value of $I_{1}$ & $I_{2}$ in equation 1

$I = \sqrt{x^{2} + 6 x + 10} - 3 [\log | (x + 3) + \sqrt{x^{2} + 6 x + 10} |] + c$

Indefinite Integrals Exercise 18.21 Question 2

Answer:

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

Given:

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

Hint: Simplify it and solve it
Solution: Let

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

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

\begin{aligned} I = \int \frac{2 x + 2}{\sqrt{x^{2} + 2 x - 1}} d x - \int \frac{1}{\sqrt{x^{2} + 2 x - 1}} d x \\ I = \int \frac{2 x + 2}{\sqrt{x^{2} + 2 x - 1}} d x - \int \frac{1}{\sqrt{x^{2} + 2 x + 1 - 1 - 1}} d x \\ I = \int \frac{2 x + 2}{\sqrt{x^{2} + 2 x - 1}} d x - \int \frac{1}{\sqrt{(x + 1)^{2} - 2}} d x \end{aligned}

Let

x^{2} + 2 x - 1 = y

d x (2 x + 2) = d y

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

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

\begin{aligned} [\int \frac{d x}{\sqrt{x^{2} - a^{2}}} = \log | x + \sqrt{x^{2} - a^{2}} | + c] \\ I = 2 \sqrt{x^{2} + 2 x - 1} - \log | x + 1 + \sqrt{x^{2} + 2 x - 1} | + c \end{aligned}

Indefinite Integrals Exercise 18.21 Question 3

Answer:

- \sqrt{4 + 5 x - x^{2}} + \frac{7}{2} \sin^{- 1} (\frac{2 x - 5}{\sqrt{41}}) + c

Given: $\int \frac{x + 1}{\sqrt{- x^{2} + 5 x + 4}} d x$
Hint: Simplify it and solve it
Solution: $I = \int \frac{x + 1}{\sqrt{- x^{2} + 5 x + 4}} d x$
$I = \int \frac{x + 1}{\sqrt{4 - (x^{2} - 5 x)}} d x = \int \frac{x + 1}{\sqrt{4 + {(\frac{5}{2})}^{2} - (x^{2} - 5 x + ({(\frac{5}{2})}^{2})}} d x$
$I = \int \frac{x + 1}{\sqrt{\frac{41}{4} - {(x - \frac{5}{2})}^{2}}} d x$
$I = \frac{1}{2} \int \frac{2 x - 5 + 7}{\sqrt{\frac{41}{4} - {(x - \frac{5}{2})}^{2}}} d x$
$= \frac{1}{2} \int \frac{2 x - 5}{\sqrt{\frac{41}{4} - {(x - \frac{5}{2})}^{2}}} d x + \frac{7}{2} \int \frac{d x}{\sqrt{\frac{41}{4} - {(x - \frac{5}{2})}^{2}}}$
$\frac{41}{4} - {(x - \frac{5}{2})}^{2} = y$
$\begin{aligned} - 2 (x - \frac{5}{2}) d x = d y \\ (2 x - 5) d x = - d y \end{aligned}$
$\Rightarrow I = \frac{1}{2} \int \frac{- d y}{\sqrt{y}} + \frac{7}{2} \int \frac{d x}{\sqrt{\frac{41}{4} - {(x - \frac{5}{2})}^{2}}}$
$\Rightarrow I = - \frac{1}{2} [\frac{(y)^{\frac{1}{2}}}{\frac{1}{2}}] + \frac{7}{2} (\sin^{- 1} (\frac{x - \frac{5}{2}}{\frac{\sqrt{41}}{2}})) + c [\int \frac{1}{\sqrt{a^{2} - x^{2}}} = \sin^{- 1} (\frac{x}{a}) + c]$
$\begin{aligned} \Rightarrow I = - {[\frac{41}{4} - {(x - \frac{5}{2})}^{2}]}^{\frac{1}{2}} + \frac{7}{2} \sin^{- 1} (\frac{2 x - 5}{\sqrt{41}}) + c \\ \Rightarrow I = - \sqrt{5 x + 4 - x^{2}} + \frac{7}{2} \sin^{- 1} (\frac{2 x - 5}{\sqrt{41}}) + c \end{aligned}$

Indefinite Integrals Exercise 18.21 Question 4

Answer: $2 \sqrt{3 x^{2} - 5 x + 1} + c$
Given: $\int \frac{6 x - 5}{\sqrt{3 x^{2} - 5 x + 1}} d x$
Hint: Simplify it and solve it
Solution:
Let

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

Let

3 x^{2} - 5 x + 1 = y

\Rightarrow (6 x - 5) d x = d y

I = \int \frac{d y}{\sqrt{y}}

I = \frac{\sqrt{y}}{\frac{1}{2}} + c

\begin{aligned} I = 2 \sqrt{y} + c \\ I = 2 \sqrt{3 x^{2} - 5 x + 1} + c \end{aligned}

(From equation (1))

Indefinite Integrals Exercise 18.21 Question 5

Answer: $- 3 \sqrt{5 - x^{2} - 2 x} - 2 \sin^{- 1} (\frac{x + 1}{\sqrt{6}}) + c$
Given: $\int \frac{3 x + 1}{\sqrt{5 - x^{2} - 2 x}} d x$
Hint: Simplify it and then integrate
Solution:
$\begin{aligned} I = \int \frac{3 x + 1}{\sqrt{5 - x^{2} - 2 x}} d x \\ I = \int \frac{3 x + 1}{\sqrt{6 - (x^{2} + 2 x + 1)}} d x \\ I = \int \frac{3 x + 1}{\sqrt{6 - (x + 1)^{2}}} d x \\ I = 3 \int \frac{x + \frac{1}{3}}{\sqrt{6 - (x + 1)^{2}}} \end{aligned}$
$= - \frac{3}{2} \int \frac{- 2 x - \frac{2}{3}}{\sqrt{6 - (x + 1)^{2}}}$ $(multiplying and dividing numerator by-2$
$\begin{aligned} I = - \frac{3}{2} \int \frac{- 2 x - 2 - \frac{2}{3} + 2}{\sqrt{6 - (x + 1)^{2}}} d x \\ I = - \frac{3}{2} \int \frac{- 2 x - 2}{\sqrt{5 - x^{2} - 2 x}} + - \frac{3}{2} (\frac{4}{3}) \int \frac{1}{\sqrt{6 - (x + 1)^{2}}} d x \end{aligned}$
$I = I_{1} + I_{2}$ .................(1)
$I_{1} = - \frac{3}{2} \int \frac{- 2 x - 2}{\sqrt{5 - x^{2} - 2 x}} d x; I_{2} = - 2 \int \frac{1}{\sqrt{6 - (x + 1)^{2}}} d x$
Now,
Let

5 - x^{2} - 2 x = y

..........(2)

(- 2 x - 2) d x = d y

\begin{aligned} \Rightarrow I_{1} = - \frac{3}{2} \int \frac{d x}{\sqrt{y}} \\ I_{1} = - \frac{3}{2} (\frac{\sqrt{y}}{\frac{1}{2}}) + c \end{aligned}

From Equation (2)

I_{1} = - 3 \sqrt{5 - x^{2} - 2 x} + c

Now,

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

I_{2} = - 2 \sin^{- 1} (\frac{x + 1}{\sqrt{6}}) + c

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

Putting value of

I_{1}

I_{2}

in equation (1) and we get

I = - 3 \sqrt{5 - x^{2} - 2 x} - 2 \sin^{- 1} (\frac{x + 1}{\sqrt{6}}) + c

Indefinite Integrals Exercise 18.21 Question 6

Answer: $- \sqrt{8 + x - x^{2}} + \frac{1}{2} \sin^{- 1} (\frac{2 x - 1}{\sqrt{33}}) + c$
Given: $\int \frac{x}{\sqrt{8 + x - x^{2}}} d x$
Hint: Simplify the given function
Solution:
$\begin{aligned} I = \int \frac{x}{\sqrt{8 + x - x^{2}}} d x \\ I = - \frac{1}{2} \int \frac{- 2 x + 1 - 1}{\sqrt{8 + x - x^{2}}} d x \\ I = - \frac{1}{2} \int \frac{- 2 x + 1}{\sqrt{8 + x - x^{2}}} d x + \frac{1}{2} \int \frac{1}{\sqrt{8 + x - x^{2}}} d x \end{aligned}$
$I = I_{1} + I_{2}$ ...................(1)
$I_{1} = - \frac{1}{2} \int \frac{- 2 x + 1}{\sqrt{8 + x - x^{2}}} d x and I_{2} = \frac{1}{2} \int \frac{1}{\sqrt{8 + x - x^{2}}} d x$
$I_{1} = - \frac{1}{2} \int \frac{- 2 x + 1}{\sqrt{8 + x - x^{2}}} d x$
Let

8 + x - x^{2} = y

.............(2)

(1 - 2 x) d x = d y

I_{1} = - \frac{1}{2} \int \frac{d y}{\sqrt{y}} = - \frac{1}{2} (\frac{\sqrt{y}}{\frac{1}{2}}) + c

\begin{aligned} I_{1} = - \sqrt{y} + c \\ I_{1} = - \sqrt{8 + x - x^{2}} + c \end{aligned}

(From Equation 2)
Now,

I_{2} = \frac{1}{2} \int \frac{1}{\sqrt{8 + x - x^{2}}} d x

I_{2} = \frac{1}{2} \int \frac{1}{\sqrt{8 - (x^{2} - x + \frac{1}{4}) + \frac{1}{4}}} d x

I_{2} = \frac{1}{2} \int \frac{d x}{\sqrt{\frac{33}{4} - {(x - \frac{1}{2})}^{2}}}

I_{2} = \frac{1}{2} \sin^{- 1} (\frac{2 x - 1}{\sqrt{33}}) + c

Putting

I_{1}

I_{2}

in equation (1)

I = - \sqrt{8 + x - x^{2}} + \frac{1}{2} \sin^{- 1} (\frac{2 x - 1}{\sqrt{33}}) + c

Indefinite Integrals Exercise 18.21 Question 7

Answer: $\sqrt{x^{2} + 2 x - 1} + \log | x + 1 + \sqrt{x^{2} + 2 x - 1} | + c$
Given: $\int \frac{x + 2}{\sqrt{x^{2} + 2 x - 1}} d x$
Hint: Simplify the given function
Solution:
$\begin{aligned} I = \int \frac{x + 2}{\sqrt{x^{2} + 2 x - 1}} d x \\ I = \frac{1}{2} \int \frac{2 x + 4}{\sqrt{x^{2} + 2 x - 1}} d x \\ I = \frac{1}{2} \int \frac{2 x + 2}{\sqrt{x^{2} + 2 x - 1}} d x + \frac{2}{2} \int \frac{1}{\sqrt{x^{2} + 2 x - 1}} d x \\ I = \frac{1}{2} \int \frac{2 x + 2}{\sqrt{x^{2} + 2 x - 1}} d x + 1 \int \frac{1}{\sqrt{x^{2} + 2 x + 1 - 2}} d x \\ I = \frac{1}{2} \int \frac{2 x + 2}{\sqrt{x^{2} + 2 x - 1}} d x + \int \frac{1}{\sqrt{(x + 1)^{2} - (\sqrt{2})^{2}}} d x \end{aligned}$
$I = \frac{1}{2} [\frac{\sqrt{x^{2} + 2 x - 1}}{\frac{1}{2}}] + \log | x + 1 + \sqrt{x^{2} + 2 x - 1} | + c$
$[\begin{array}{l} U \sin g \\ (f (x))^{n} f^{1} (x) d x = \frac{[f (x)]^{n + 1}}{n + 1} \\ \int \frac{1}{\sqrt{x^{2} - a^{2}}} d x = \log | x + \sqrt{x^{2} - a^{2}} | + c \end{array}]$
$I = \sqrt{x^{2} + 2 x - 1} + \log | x + 1 + \sqrt{x^{2} + 2 x - 1} | + c$

Indefinite Integrals Exercise 18.21 Question 8

Answer: $\sqrt{x^{2} - 1} + 2 \log | x + \sqrt{x^{2} - 1} | + c$
Given: $\int \frac{x + 2}{\sqrt{x^{2} - 1}} d x$
Hint: Simplify the given function
Solution: $I = \int \frac{x + 2}{\sqrt{x^{2} - 1}} d x$
$\begin{aligned} I = \frac{1}{2} \int \frac{2 x}{\sqrt{x^{2} - 1}} d x + 2 \int \frac{1}{\sqrt{x^{2} - 1}} d x \\ I = \frac{1}{2} [\frac{x^{2} - 1}{\frac{1}{2}}] + 2 \log | x + \sqrt{x^{2} - 1} | + c \end{aligned}$
$[\begin{array}{l} U \sin g \\ \int (f (x))^{n} f^{1} (x) d x = \frac{f (x)^{n + 1}}{n + 1} + c \\ \int \frac{1}{\sqrt{x^{2} - a^{2}}} d x = \log | x + \sqrt{x^{2} - a^{2}} | + c \end{array}]$
$I = \sqrt{x^{2} - 1} + 2 \log | x + \sqrt{x^{2} - 1} | + c$

Indefinite Integrals Exercise 18.21 Question 9

Answer: $\sqrt{x^{2} + 1} - \log | x + \sqrt{x^{2} + 1} | + c$
Given: $\int \frac{x - 1}{\sqrt{x^{2} + 1}} d x$
Hint: Simplify the given function
Solution:
$\begin{aligned} I = \int \frac{x - 1}{\sqrt{x^{2} + 1}} d x \\ I = \frac{1}{2} \int \frac{2 x}{\sqrt{x^{2} + 1}} d x - \int \frac{1}{\sqrt{x^{2} + 1}} d x \\ I = \frac{1}{2} [\frac{\sqrt{x^{2} + 1}}{\frac{1}{2}}] - \log | x + \sqrt{x^{2} + 1} | + c \end{aligned}$
$[\begin{array}{l} U \sin g \\ \int (f (x))^{n} f^{1} (x) d x = \frac{[f (x)]^{n + 1}}{n + 1} + c \\ \int \frac{1}{\sqrt{x^{2} + a^{2}}} d x = \log | x + \sqrt{x^{2} + a^{2}} | + c \end{array}]$
$I = \sqrt{x^{2} + 1} - \log | x + \sqrt{x^{2} + 1} | + c$

Indefinite Integrals Exercise 18.21 Question 10

Answer: $\sqrt{x^{2} + x + 1} - \frac{1}{2} \log | \frac{2 x + 1}{2} + \sqrt{x^{2} + x + 1} | + c$
Given: $\int \frac{x}{\sqrt{x^{2} + x + 1}} d x$
Hint: Simplify the given

(f (x))^{n}

Solution:
$\begin{aligned} I = \int \frac{x}{\sqrt{x^{2} + x + 1}} d x \\ I = \frac{1}{2} \int \frac{2 x}{\sqrt{x^{2} + x + 1}} d x \\ I = \frac{1}{2} \int \frac{2 x + 1 - 1}{\sqrt{x^{2} + x + 1}} d x \\ I = \frac{1}{2} \int \frac{2 x + 1}{\sqrt{x^{2} + x + 1}} d x - \frac{1}{2} \int \frac{1}{\sqrt{x^{2} + x + 1}} d x \\ I = \frac{1}{2} \int \frac{2 x + 1}{\sqrt{x^{2} + x + 1}} d x - \frac{1}{2} \int \frac{1}{\sqrt{(x^{2} + x + \frac{1}{4}) + 1 - \frac{1}{4}}} d x \end{aligned}$
$\begin{aligned} I = \frac{1}{2} \int \frac{2 x + 1}{\sqrt{x^{2} + x + 1}} d x - \frac{1}{2} \int \frac{1}{\sqrt{{(x + \frac{1}{2})}^{2} + \frac{3}{4}}} d x \\ I = \frac{1}{2} [\frac{\sqrt{x^{2} + x + 1}}{\frac{1}{2}}] - \frac{1}{2} \log | x + \frac{1}{2} + \sqrt{x^{2} + x + 1} | + c \end{aligned}$
$[\begin{array}{l} U \sin g \\ \int (f (x))^{n} f^{1} (x) d x = \frac{[f (x)]^{n + 1}}{n + 1} + c \\ \int \frac{1}{\sqrt{x^{2} + a^{2}}} d x = \log | x + \sqrt{x^{2} + a^{2}} | + c \end{array}]$
$I = \sqrt{x^{2} + x + 1} - \frac{1}{2} \log | \frac{2 x + 1}{2} + \sqrt{x^{2} + x + 1} | + c$

Indefinite Integrals Exercise 18.21 Question 11

Answer: $\sqrt{x^{2} + 1} + \log | x + \sqrt{x^{2} + 1} | + c$
Given: $\int \frac{x + 1}{\sqrt{x^{2} + 1}} d x$
Hint: Simplify the given integration and solve it
Solution:
$\begin{aligned} I = \int \frac{x + 1}{\sqrt{x^{2} + 1}} d x \\ I = \int \frac{x}{\sqrt{x^{2} + 1}} d x + \int \frac{1}{\sqrt{x^{2} + 1}} d x \end{aligned}$
$\begin{aligned} I = \frac{1}{2} \int \frac{2 x}{\sqrt{x^{2} + 1}} d x + \log | x + \sqrt{x^{2} + 1} | + c \\ I = I_{1} + \log | x + \sqrt{x^{2} + 1} | + c \\ I_{1} = \frac{1}{2} \int \frac{2 x}{\sqrt{x^{2} + 1}} d x \end{aligned}$
$\begin{aligned} Let \\ x^{2} + 1 = y \\ 2 x d x = d y \end{aligned}$
$\begin{aligned} I_{1} = \frac{1}{2} \int \frac{d y}{\sqrt{y}} = \frac{1}{2} (\frac{\sqrt{y}}{\frac{1}{2}}) + c \\ I_{1} = \sqrt{y} + c \\ I_{1} = \sqrt{x^{2} + 1} + c \\ I = \sqrt{x^{2} + 1} + \log | x + \sqrt{x^{2} + 1} | + c \end{aligned}$

Indefinite Integrals Exercise 18.21 Question 12

Answer: $2 \sqrt{x^{2} + 2 x + 5} + 3 \log | x + 1 + \sqrt{x^{2} + 2 x + 5} | + c$
Given: $\int \frac{2 x + 5}{\sqrt{x^{2} + 2 x + 5}} d x$

Hint: Simplify the given (f(x))ⁿ
Solution:
$\begin{aligned} I = \int \frac{2 x + 5}{\sqrt{x^{2} + 2 x + 5}} d x \\ I = \int \frac{2 x + 2 + 3}{\sqrt{x^{2} + 2 x + 5}} d x \\ I = \int \frac{2 x + 2}{\sqrt{x^{2} + 2 x + 5}} d x + 3 \int \frac{1}{\sqrt{x^{2} + 2 x + 5}} d x \end{aligned}$
$I = I_{1} + I_{2}$ .......................(1)
Where
$I_{1} = \int \frac{2 x + 2}{\sqrt{x^{2} + 2 x + 5}} d x & I_{2} = 3 \int \frac{1}{\sqrt{x^{2} + 2 x + 5}} d x$
Now,Let
$\begin{aligned} x^{2} + 2 x + 5 = y \\ (2 x + 2) d x = d y \end{aligned}$ ..........................(2)
$\begin{aligned} I_{1} = 2 \sqrt{y} + c \\ I_{1} = 2 \sqrt{x^{2} + 2 x + 5} + c \end{aligned}$ ( From equation 2)
Now, $I_{2} = 3 \int \frac{1}{\sqrt{x^{2} + 2 x + 4 + 1}} d x$
$I_{2} = 3 \int \frac{1}{\sqrt{(x + 1)^{2} + 2^{2}}} d x$ $[\int \frac{1}{\sqrt{x^{2} + a^{2}}} d x = \log | x + \sqrt{x^{2} + a^{2}} | + c]$
$I_{2} = 3 \log | x + 1 + \sqrt{x^{2} + 2 x + 5} | + c$
Putting value of $I_{1}$ & $I_{2}$ in equation (1)
$I = 2 \sqrt{x^{2} + 2 x + 5} + 3 \log | x + 1 + \sqrt{x^{2} + 2 x + 5} | + c$

Indefinite Integrals Exercise 18.21 Question 13

Answer: $- 3 \sqrt{5 - x^{2} - 2 x} - 2 \sin^{- 1} (\frac{x + 1}{\sqrt{6}}) + c$
Given: $\int \frac{3 x + 1}{\sqrt{5 - x^{2} - 2 x}} d x$
Hint: Simplify the given Function
Solution:
$\begin{aligned} I = \int \frac{3 x + 1}{\sqrt{5 - x^{2} - 2 x}} d x \\ I = 3 \int \frac{x + \frac{1}{3}}{\sqrt{5 - x^{2} - 2 x}} d x \end{aligned}$
Multiplying and dividing the numerator by -2,

\begin{aligned} I & = \frac{- 3}{2} \int \frac{- 2 x - \frac{2}{3}}{\sqrt{5 - x^{2} - 2 x}} d x \\ I & = \frac{- 3}{2} \int \frac{- 2 x - 2 + \frac{4}{3}}{\sqrt{5 - x^{2} - 2 x}} d x \\ I & = \frac{- 3}{2} \int \frac{- 2 x - 2}{\sqrt{5 - x^{2} - 2 x}} d x - 2 \int \frac{1}{\sqrt{5 - x^{2} - 2 x}} d x \\ I & = I_{1} + I_{2} \\ I_{1} & = \frac{- 3}{2} \int \frac{- 2 x - 2}{\sqrt{5 - x^{2} - 2 x}} d x; I_{2} = - 2 \int \frac{1}{\sqrt{5 - x^{2} - 2 x}} d x \end{aligned}

Now,

I_{1} = \frac{- 3}{2} \int \frac{- 2 x - 2}{\sqrt{5 - x^{2} - 2 x}} d x

Let

\begin{aligned} 5 - x^{2} - 2 x = y \\ (- 2 - 2 x) d x = d y \end{aligned}

\Rightarrow I_{1} = \frac{- 3}{2} \int \frac{d y}{\sqrt{y}} = \frac{- 3}{2} (\frac{\sqrt{y}}{\frac{1}{2}}) + c

\begin{aligned} I_{1} = - 3 \sqrt{y} + c \\ I_{1} = - 3 \sqrt{5 - x^{2} - 2 x} + c \end{aligned}

Now,

I_{2} = - 2 \int \frac{1}{\sqrt{5 - x^{2} - 2 x}} d x

\begin{aligned} I_{2} = - 2 \int \frac{1}{\sqrt{5 - (x^{2} + 2 x + 1) + 1}} d x \\ I_{2} = - 2 \int \frac{1}{\sqrt{6 - (x + 1)^{2}}} d x \\ I_{2} = - 2 \sin^{- 1} [\frac{x + 1}{\sqrt{6}}] + c \end{aligned}

Putting

I_{1}

I_{2}

in equation (1) we get

I = - 3 \sqrt{5 - x^{2} - 2 x} - 2 \sin^{- 1} (\frac{x + 1}{\sqrt{6}}) + c

Indefinite Integrals Exercise 18.21 Question 14

Answer: $\sin^{- 1} x + \sqrt{1 - x^{2}} + c$
Given: $\int \sqrt{\frac{1 - x}{1 + x}} d x$
Hint: Simplify the given function
Solution:
$\begin{aligned} I & = \int \sqrt{\frac{1 - x}{1 + x}} d x \\ I & = \int \sqrt{\frac{1 - x}{1 + x} \times \frac{1 - x}{1 - x}} d x (rationalizing ta e denominator) \\ = \int \sqrt{\frac{(1 - x)^{2}}{1 - x^{2}}} d x [(a + b) (a - b) = (a^{2} - b^{2})] \\ I & = \int \frac{1 - x}{\sqrt{1 - x^{2}}} d x \end{aligned}$
$\begin{aligned} I = \int \frac{1}{\sqrt{1 - x^{2}}} d x - \frac{1}{2} \int \frac{2 x}{\sqrt{1 - x^{2}}} d x \\ I = \sin^{- 1} x + \frac{1}{2} (\frac{\sqrt{1 - x^{2}}}{\frac{1}{2}}) + c \end{aligned}$
$\begin{aligned} [\begin{array}{c} \int \frac{d x}{\sqrt{a^{2} - x^{2}}} = \sin^{- 1} (\frac{x}{a}) + c \\ \int (f (x))^{n} f^{1} (x) d x = \frac{[f (x)]^{n + 1}}{n + 1} + c \end{array}] \\ I = \sin^{- 1} x + \sqrt{1 - x^{2}} + c \end{aligned}$

Indefinite Integrals Exercise 18.21 Question 15

Answer: $2 \sqrt{x^{2} + 4 x + 3} - 3 \log | x + 2 + \sqrt{x^{2} + 4 x + 3} | + c$
Given: $\int \frac{2 x + 1}{\sqrt{x^{2} + 4 x + 3}} d x$
Hint: Simplify the given function
Solution:
$\begin{aligned} I = \int \frac{2 x + 1}{\sqrt{x^{2} + 4 x + 3}} d x \\ I = \int \frac{2 x + 4 - 3}{\sqrt{x^{2} + 4 x + 3}} d x \\ I = \int \frac{2 x + 4}{\sqrt{x^{2} + 4 x + 3}} d x - \int \frac{3}{\sqrt{x^{2} + 4 x + 3}} d x \\ I = \int \frac{2 x + 4}{\sqrt{x^{2} + 4 x + 3}} d x - \int \frac{3}{\sqrt{x^{2} + 4 x + 4 - 4 + 3}} d x \\ I = \int \frac{2 x + 4}{\sqrt{x^{2} + 4 x + 3}} d x - \int \frac{3}{\sqrt{(x + 2)^{2} - 1}} d x \end{aligned}$
$\begin{aligned} I = I_{1} - 3 \log | x + 2 + \sqrt{x^{2} + 4 x + 3} | + c \dots [\int \frac{1}{\sqrt{x^{2} - a^{2}}} d x = \log | x + \sqrt{x^{2} - a^{2}} | + c] \\ I_{1} = \int \frac{2 x + 4}{\sqrt{x^{2} + 4 x + 3}} d x \end{aligned}$
Let,
$\begin{aligned} x^{2} + 4 x + 3 = y \\ (2 x + 4) d x = d y \\ I_{1} = \int \frac{d y}{\sqrt{y}} = \frac{\sqrt{y}}{\frac{1}{2}} + c \\ I_{1} = 2 \sqrt{x^{2} + 4 x + 3} + c \\ I = 2 \sqrt{x^{2} + 4 x + 3} - 3 \log | x + 2 + \sqrt{x^{2} + 4 x + 3} | + c \end{aligned}$

Indefinite Integrals Exercise 18.21 Question 16

Answer: $2 \sqrt{x^{2} + 4 x + 5} - \log | x + 2 + \sqrt{x^{2} + 4 x + 5} | + c$
Given: $\int \frac{2 x + 3}{\sqrt{x^{2} + 4 x + 5}} d x$
Hint: Simplify the given function
Solution:
$\begin{aligned} I = \int \frac{2 x + 3}{\sqrt{x^{2} + 4 x + 5}} d x \\ I = \int \frac{2 x + 4 - 1}{\sqrt{x^{2} + 4 x + 5}} d x \\ I = \int \frac{2 x + 4}{\sqrt{x^{2} + 4 x + 5}} d x - \int \frac{1}{\sqrt{x^{2} + 4 x + 4 - 4 + 5}} d x \\ I = \int \frac{2 x + 4}{\sqrt{x^{2} + 4 x + 5}} d x - \int \frac{d x}{\sqrt{(x + 2)^{2} + 1}} \\ I = \int \frac{2 x + 4}{\sqrt{x^{2} + 4 x + 5}} d x - \log | x + 2 + \sqrt{x^{2} + 4 x + 5} | + c \end{aligned}$
Using.................... $[\int \frac{1}{\sqrt{x^{2} + a^{2}}} d x = \log | x + \sqrt{x^{2} + a^{2}} | + c]$
Let

\begin{aligned} x^{2} + 4 x + 5 = y \\ (2 x + 4) d x = d y \end{aligned}

\Rightarrow \int \frac{2 x + 4}{\sqrt{x^{2} + 4 x + 5}} d x = \int \frac{d y}{\sqrt{y}} = \frac{\sqrt{y}}{\frac{1}{2}} + c

\begin{aligned} = 2 \sqrt{y} + c \\ = 2 \sqrt{x^{2} + 4 x + 5} + c \end{aligned}

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

Indefinite Integrals Exercise 18.21 Question 17

Answer: $5 \sqrt{x^{2} + 4 x + 10} - 7 \log | x + 2 + \sqrt{x^{2} + 4 x + 10} |$
Given: $\int \frac{5 x + 3}{\sqrt{x^{2} + 4 x + 10}} d x$
Hint: Simplify the given function
Solution:
$\begin{aligned} I = \int \frac{5 x + 3}{\sqrt{x^{2} + 4 x + 10}} d x \\ I = 5 \int \frac{x + \frac{3}{5}}{\sqrt{x^{2} + 4 x + 10}} d x multiplying and dividing by 2 in numerator, \\ I = \frac{5}{2} \int \frac{2 x + \frac{6}{5}}{\sqrt{x^{2} + 4 x + 10}} d x \\ I = \frac{5}{2} \int \frac{2 x + 4 - \frac{14}{5}}{\sqrt{x^{2} + 4 x + 10}} d x \end{aligned}$
$\begin{aligned} I = \frac{5}{2} \int \frac{2 x + 4}{\sqrt{x^{2} + 4 x + 10}} d x - 7 \int \frac{1}{\sqrt{x^{2} + 4 x + 10}} d x \\ I = \frac{5}{2} \int \frac{2 x + 4}{\sqrt{x^{2} + 4 x + 10}} d x - 7 \int \frac{1}{\sqrt{x^{2} + 4 x + 4 + 6}} d x \\ I = \frac{5}{2} \int \frac{2 x + 4}{\sqrt{x^{2} + 4 x + 10}} d x - 7 \int \frac{1}{\sqrt{(x + 2)^{2} + (\sqrt{6})^{2}}} d x \\ I = \frac{5}{2} \int \frac{2 x + 4}{\sqrt{x^{2} + 4 x + 10}} d x - 7 \log | x + 2 + \sqrt{x^{2} + 4 x + 10} | \end{aligned}$
using....................

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

Let

\begin{aligned} x^{2} + 4 x + 10 = y \\ (2 x + 4) d x = d y \end{aligned}

\Rightarrow \int \frac{2 x + 4}{\sqrt{x^{2} + 4 x + 10}} d x = \int \frac{d y}{\sqrt{y}} = \frac{\sqrt{y}}{\frac{1}{2}} + c

\begin{aligned} = 2 \sqrt{y} + c \\ = 2 \sqrt{x^{2} + 4 x + 10} + c \end{aligned}

I = 5 \sqrt{x^{2} + 4 x + 10} - 7 \log | x + 2 + \sqrt{x^{2} + 4 x + 10} |

Indefinite Integrals Exercise 18.21 Question 18

Answer: $\sqrt{x^{2} + 2 x + 3} + \log | x + 1 + \sqrt{x^{2} + 2 x + 3} | + c$
Given: $\int \frac{x + 2}{\sqrt{x^{2} + 2 x + 3}} d x$
Hint: Simplify the given function
Solution:
$\begin{aligned} I = \int \frac{x + 2}{\sqrt{x^{2} + 2 x + 3}} d x \\ I = \frac{1}{2} \int \frac{2 x + 4}{\sqrt{x^{2} + 2 x + 3}} d x \\ I = \frac{1}{2} \int \frac{2 x + 2}{\sqrt{x^{2} + 2 x + 3}} d x + \frac{1}{2} \int \frac{2}{\sqrt{x^{2} + 2 x + 3}} d x \\ I = \frac{1}{2} \int \frac{2 x + 2}{\sqrt{x^{2} + 2 x + 3}} d x + \int \frac{1}{\sqrt{x^{2} + 2 x + 1 + 2}} d x \\ I = \frac{1}{2} \int \frac{2 x + 2}{\sqrt{x^{2} + 2 x + 3}} d x + \int \frac{1}{\sqrt{(x + 1)^{2} + 2}} d x \end{aligned}$
$I = \frac{1}{2} \int \frac{2 x + 2}{\sqrt{x^{2} + 2 x + 3}} d x + \log | (x + 1) + \sqrt{x^{2} + 2 x + 3} | + c | \int \frac{1}{\sqrt{x^{2} + a^{2}}} d x = \log | x + \sqrt{x^{2} + a^{2}} | + c |$
Let

\begin{aligned} x^{2} + 2 x + 3 = y \\ (2 x + 2) d x = d y \end{aligned}

\Rightarrow \int \frac{2 x + 2}{\sqrt{x^{2} + 2 x + 3}} d x = \int \frac{d y}{\sqrt{y}} = \frac{\sqrt{y}}{\frac{1}{2}} + c

\begin{aligned} = 2 \sqrt{y} + c \\ = 2 \sqrt{x^{2} + 2 x + 3} + c \end{aligned}

\begin{aligned} \Rightarrow I = \frac{1}{2} (2 \sqrt{x^{2} + 2 x + 3}) + \log | x + 1 + \sqrt{x^{2} + 2 x + 3} | + c \\ \Rightarrow I = \sqrt{x^{2} + 2 x + 3} + \log | x + 1 + \sqrt{x^{2} + 2 x + 3} | + c \end{aligned}

Rd Sharma Class 12th Exercise 18.21 solutions will help every student because of the plenty of benefits. Some benefits of these solutions are listed below:

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Frequently Asked Questions (FAQs)

1. How can RD Sharma class 12 solutions help with Maths?

The key feature of these solutions is that each question is followed by a step-by-step explanation. There are no missed steps or shortcuts. As a result, if the student pays close attention, the answer will become self-explanatory. It also aids students in solidifying their concepts and acing their exams.

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Yes, having all of your answers in one place is useful for last-minute revision so you don't have to go looking for solutions. RD Sharma class 11 solutions for Mathematics will collect all of the answers to the exercises in one location for easy reference.

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Yes, it can be a valuable resource for exam preparation and can help you achieve a high score. Solving more and more textbook solutions gives you enough practice.

5. How do you find the indefinite integral's value?

The process of finding the indefinite integral value of a function is also called integration or integrating f(x). This can be presented as:

∫f(x)dx = F(x) + C, where C is any real number.

We generally use suitable formulas that help in getting the antiderivative of the given function.

The result of the indefinite integral is a function.

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