RD Sharma Class 12 Exercise 18.32 Indefinite Integrals Solutions Maths

RD Sharma Class 12 Exercise 18.32 Indefinite Integrals Solutions Maths - Download PDF Free Online

Edited By Kuldeep Maurya | Updated on Jan 24, 2022 10:54 AM IST

RD Sharma books are considered as some of the best materials to study for exams. Majority of CBSE schools and teachers follow this book for their lectures as well as the set of question papers. This is why it is a leading choice among students as it is informative and detailed. RD Sharma Solutions Indefinite Integrals Ex 18.32 material is prepared by a group of experts to help students get a better understanding of the subject. The Class 12 RD Sharma chapter 18 exercise 18.32 solution meets a fantastic essential of maths questions. It contains step-by-step solutions for each answer making it easier for students to understand.

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RD Sharma Class 12 Solutions Chapter18 Indefinite Integrals - Other Exercise

Indefinite Integrals Excercise:18.32

Indefinite Integrals Excercise 18.32 Question 2

Answer : -

\frac{1}{\sqrt{5}} \log | \frac{\sqrt{2 x + 3} - \sqrt{5}}{\sqrt{2 x + 3} + \sqrt{5}} | + C

Hint :- Use substitution method to solve this integral.
Given :-

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

Sol : - Let

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

Put

2 x + 3 = t^{2} \Rightarrow 2 d x = 2 t d t \to d x = t d t

.then,

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

\begin{aligned} = 2 \cdot \frac{1}{2 \sqrt{5}} \log | \frac{t - \sqrt{5}}{t + \sqrt{5}} | + C (∵ \int \frac{1}{x^{2} - a^{2}} d x = 1 / 2 a \log | \frac{x - a}{x + a} | + C) \\ = \frac{1}{\sqrt{5}} \log | \frac{\sqrt{2 x + 3} - \sqrt{5}}{\sqrt{2 x + 3} + \sqrt{5}} | + C (∵ t = \sqrt{2 x + 3}) Ans.. \end{aligned}

Indefinite Integrals Excercise 18.32 Question 3

Answer : -

2 \sqrt{x + 2} + \frac{2}{\sqrt{3}} \log | \frac{\sqrt{x + 2} - \sqrt{3}}{\sqrt{x + 2} + \sqrt{3}} | + C

Hint :- Use substitution method to solve this integral.
Given :-

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

Sol : -

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

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

Now, I = I1 +I2 ........................ (I)
Where

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

Therefore,

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

Let

x + 2 = u \Rightarrow d x = du

Then,

\begin{aligned} I_{1} & = \int \frac{1}{u^{\frac{1}{2}}} d u = \int u^{\frac{- 1}{2}} du = \frac{u^{\frac{- 1}{2}} + 1}{\frac{- 1}{2} + 1} + C (∵ \int x^{n} d x = \frac{x^{n + 1}}{n + 1} + C) \\ = \frac{u^{\frac{1}{2}}}{\frac{1}{2}} + C_{1} \\ = 2 \sqrt{x + 2} + C_{1} \dots \dots \dots \dots (II) \end{aligned}

And

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

Let

x + 2 = t^{2} => d x = 2 t d t

. Then,

\begin{aligned} I_{2} & = \int \frac{2}{(t^{2} - 2 - 1) \sqrt{t^{2}}} 2 t d t = 4 \int \frac{t d t}{(t^{2} - 3) t} & (∵ t^{2} - 2 = x) \\ = 4 \int \frac{1}{(t^{2}) - \sqrt{3^{2}}} dt = \frac{4}{2 \cdot \sqrt{3}} \log | \frac{t - \sqrt{3}}{t + \sqrt{3}} | + C & (∵ \int \frac{1}{x^{2} - a^{2}} d x = 1 / 2 a \log | \frac{x - a}{x + a} | + C) \\ = \frac{2}{\sqrt{3}} \log | \frac{\sqrt{x + 2} - \sqrt{3}}{\sqrt{x + 2} + \sqrt{3}} | + C_{2} & (∵ t = \sqrt{x + 2}) \dots \dots \dots . III) \end{aligned}

Put the value of Equ.(II) and (III) in (I) then,

\begin{aligned} I = 2 \sqrt{x + 2} + C_{1} + \frac{2}{\sqrt{3}} \log | \frac{\sqrt{x + 2} - \sqrt{3}}{\sqrt{x + 2} + \sqrt{3}} | + C_{2} \\ = 2 \sqrt{x + 2} + + \frac{2}{\sqrt{3}} \log | \frac{\sqrt{x + 2} - \sqrt{3}}{\sqrt{x + 2} + \sqrt{3}} | + C (∵ C = C_{1} + C_{2}) Ans.. \end{aligned}

Indefinite Integrals Excercise 18.32 Question 4

Answer :

\frac{2}{3} (x + 2)^{\frac{3}{2}} - 2 \sqrt{x + 2} + \frac{1}{\sqrt{3}} \log | \frac{\sqrt{x + 2} - \sqrt{3}}{\sqrt{x + 2} + \sqrt{3}} | + C

Hint :- Use substitution method and special integration formula to solve this problem.
Given :- $\int \frac{x^{2}}{(x - 1) \sqrt{x + 2}} d x$
Sol :-

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

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

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

Now,

\int \sqrt{x + 2} d x

Put

x + 2 = p => d x = d p

then

\begin{aligned} \int \sqrt{x + 2} d x & = \int \sqrt{p} d p = \int p^{\frac{1}{2} d} = \frac{p \frac{1}{2} + 1}{\frac{1}{2} + 1} + C_{1} & (∵ \int x^{n} d x = \frac{x^{n + 1}}{n + 1} + C) \\ = \frac{p^{\frac{8}{2}}}{3 / 2} = 2 / 3 (x + 2)^{3 / 2} + C_{1} \dots \dots . . . (2) & (∵ p = x + 2) \end{aligned}

And,

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

Put

x + 2 = U => d x = d u

, then

\begin{aligned} ∴ \int \frac{1}{\sqrt{x + 2}} d x & = \int \frac{1}{\sqrt{u}} d u = \int u^{- 1 / 2} d u \\ = \frac{u \frac{- 1}{2} + 1}{- 1 / 2^{+ 1}} + C_{2} (∵ \int x^{n} d x = \frac{x^{n + 1}}{n + 1} + C) \\ = \frac{u \frac{- 1}{2} + 1}{- 1 / 2^{+ 1}} + C_{2} = 2 \sqrt{u} + C_{2} \\ = 2 \sqrt{x + 2} + C_{2} \dots \dots . . (III) (∵ u = x + 2) \end{aligned}

Also,

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

Put

x + 2 = t^{2} => d x = 2 t d t

, then

\begin{aligned} \int \frac{1}{(x - 1) \sqrt{x + 2}} dx & = \int \frac{1}{(t^{2} - 2 - 1) \sqrt{t^{2}}} 2 t d t \\ = 2 \int \frac{1}{(t^{2} - 3) t} t d t \\ = 2 \int \frac{1}{t^{2} - (\sqrt{3})^{2}} d t \end{aligned}

\begin{aligned} = 2 \cdot \frac{1}{2 . \sqrt{3}} \log | \frac{t - \sqrt{3}}{t + \sqrt{3}} | + C_{3} (∵ \int \frac{1}{x^{2} - a^{2}} d x = \frac{1}{2 a} \log | \frac{x - a}{x + a} | + C \\ = \frac{1}{\sqrt{3}} \log | \frac{\sqrt{x + 2} - \sqrt{3}}{\sqrt{x + 2} + \sqrt{3}} | + C_{3} \dots . (IV) \end{aligned}

Put the values of equ.(II), (III) and (IV) in (I), then,

\begin{aligned} ∴ & I = \frac{2}{3} (x + 2)^{3 / 2} + C_{1} - (2 \sqrt{x + 2} + C_{2}) + \frac{1}{\sqrt{3}} \log | \frac{\sqrt{x + 2} - \sqrt{3}}{\sqrt{x + 2} + \sqrt{3}} | + C_{1} \\ I & = \frac{2}{3} (x + 2)^{3 / 2} - 2 \sqrt{x + 2} + \frac{1}{\sqrt{3}} \log | \frac{\sqrt{x + 2} - \sqrt{3}}{\sqrt{x + 2} + \sqrt{3}} | + C & (∵ C = C_{1} - C_{2} + C_{3}) Ans.. \end{aligned}

Indefinite Integrals Excercise 18.32 Question 5

Answer : - $2 \sqrt{x + 1} + \frac{3}{2} \log | \frac{\sqrt{x + 1} - 2}{\sqrt{x + 1} + 2} | + C$
Hint :- Use substitution method and special integration formula to solve this problem..
Given :-

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

Sol : - Let

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

\begin{aligned} = \int \frac{x - 3 + 3}{(x - 3) \sqrt{x + 1}} d x = \int \frac{(x - 3) + 3}{(x - 3) \sqrt{x + 1}} d x \\ = \int {\frac{x - 3}{(x - 3) \sqrt{x + 1}} + \frac{3}{(x - 3) \sqrt{x + 1}}} d x \\ = \int {\frac{1}{\sqrt{x + 1}} + \frac{3}{(x - 3) \sqrt{x + 1}}} d x \\ = \int \frac{1}{\sqrt{x + 1}} d x + \int \frac{3}{(x - 3) \sqrt{x + 1}} d x \\ = \int \frac{1}{\sqrt{x + 1}} d x + 3 \int \frac{1}{(x - 3) \sqrt{x + 1}} d x \end{aligned}

Now,

\int \frac{1}{\sqrt{x + 1}} d x

Let

x + 1 = p => d x = d p

. Then,

\begin{aligned} ∴ \int \frac{1}{\sqrt{x + 1}} d x & = \int \frac{1}{\sqrt{p}} d p = \int p^{- 1 / 2} dp \\ = \frac{p^{- 1 / 2 + 1}}{- 1 / 2^{+ 1}} + C_{1} (∵ \int x^{n} d x = \frac{x^{n + 1}}{n + 1} + C) \\ = \frac{p^{1 / 2}}{1 / 2} + C_{1} = 2 \sqrt{x + 1} + C_{1} \dots \dots_{1} (II) (∵ p = x + 1) \end{aligned}

And,

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

Put

x + 1 = t^{2} => d x = 2 t d t

. Then,

\begin{aligned} ∴ \int \frac{1}{(x - 3) \sqrt{x + 1}} d x = & \int \frac{1}{(t^{2} - 1 - 3) \sqrt{t^{2}}} 2 t d t (∵ x = t^{2} - 1) \\ = 2 \int \frac{1}{(t^{2} - 4) t} t d t \\ = 2 \int \frac{1}{(t^{2}) - (2)^{2}} d t \end{aligned}

\begin{array}{ll} = 2 \cdot \frac{1}{2.2} \log | \frac{t - 2}{t + 2} | + C_{2} & (∵ \int \frac{1}{x^{2} - a^{2}} d x = \frac{1}{2 a} \log | \frac{x - a}{x + a} | + C) \\ = \frac{1}{2} \log | \frac{\sqrt{x + 1} - 2}{\sqrt{x + 1} + 2} | + C_{2} \dots \dots . III) & (∵ t = \sqrt{x + 1}) \end{array}

Put the values of equ. (II) and (III) in (I), Then ,

\begin{aligned} I = 2 \sqrt{x + 1} + C_{1} + 3 \cdot \frac{1}{2} \log | \frac{\sqrt{x + 1} - 2}{\sqrt{x + 1} + 2} | + C_{2} \\ I = 2 \sqrt{x + 1} + \frac{3}{2} \log | \frac{\sqrt{x + 1} - 2}{\sqrt{x + 1} + 2} | + C (∵ C = C_{1} + C_{2}) Ans. \end{aligned}

Indefinite Integrals Excercise 18.32 Question 6

Answer : -

\frac{1}{\sqrt{2}} \tan^{- 1} \frac{x - 1}{\sqrt{2 x}} - - \frac{1}{2 \sqrt{2}} \log | \frac{x + 1 - \sqrt{2 x}}{x + 1 + \sqrt{2 x}} | + C

Hint :- Use substitution method and special integration formula.
Given :-

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

Sol : - Let

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

Put

x = t^{2} => d x = 2 t d t

. then,

\begin{aligned} I & = \int \frac{1}{({(t^{2})}^{2} + 1) \sqrt{t^{2}}} 2 t d t \\ = \int \frac{1}{(t^{4} + 1) t} 2 t d t \end{aligned}

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

(? dividing numerator and denominator by t2)

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

\begin{aligned} = \int \frac{(1 + \frac{1}{t^{2}})}{t^{2} + \frac{1}{t^{2}}} d t - \int \frac{(1 - \frac{1}{t^{2}})}{t^{2} + \frac{1}{t^{2}}} d t \\ = \int \frac{(1 + \frac{1}{t^{2}})}{t^{2} + \frac{1}{t^{2}} + 2 - 2} d t - \int \frac{(1 - \frac{1}{t^{2}})}{t^{2} + \frac{1}{t^{2}} + 2 - 2} d t \\ \begin{aligned} = \int \frac{(1 + \frac{1}{+ t^{2}})}{{(t - \frac{1}{t})}^{2}) + 2} d t - \int \frac{(1 - \frac{1}{t^{2}})}{{(t + \frac{1}{t})}^{2}) - 2} d t \dots \dots (I) (∵ (a + b)^{2} = a^{2} + 2 a b + b^{2} \\ (a - b)^{2} = a^{2} - 2 a b + b^{2}) \end{aligned} \end{aligned}

\begin{aligned} Now, \int \frac{(1 + \frac{1}{t^{2}})}{{(t - \frac{1}{t})}^{2} + 2)} d t \\ Put t - 1 / t = u => 1 - (\frac{- 1}{t^{2}}) dt = d u \\ \Rightarrow (1 + \frac{1}{t^{2}}) d t = d u then \\ ∴ \int \frac{(1 + \frac{1}{t^{2}})}{{(t - \frac{1}{t})}^{2} + 2)} d t = \int \frac{d u}{(u)^{2} + 2)} = \int \frac{d u}{(u)^{2} + (\sqrt{2})^{2})} \end{aligned}

\begin{aligned} = \frac{1}{\sqrt{2}} \tan^{- 1} (\frac{u}{\sqrt{2}}) + C_{1} (∵ \int \frac{d x}{a^{2} + x^{2}} = \frac{1}{a} \tan^{- 1} (\frac{x}{a}) + C) \\ = \frac{1}{\sqrt{2}} \tan^{- 1} \frac{t - \frac{1}{t}}{\sqrt{2}} + C_{1} (∵ u = t - 1 / t) \\ = \frac{1}{\sqrt{2}} \tan^{- 1} \frac{t^{2} - 1}{t} + C_{1} \\ = \frac{1}{\sqrt{2}} \tan^{- 1} \frac{t^{2} - 1}{\sqrt{2 t}} + C_{1} \dots \dots . . . (II) \end{aligned}

And

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

Put

t + 1 / t = v => (1 - 1 / t^{2}) dt = d y

then,

\begin{aligned} ∴ \int \frac{(1 - \frac{1}{t^{2}})}{{(t + \frac{1}{t})}^{2} - 2)} d t & = \int \frac{d v}{(v)^{2} + 2)} = \int \frac{d v}{(v)^{2} + (\sqrt{2})^{2})} \\ = \frac{1}{2 . \sqrt{2}} \log | \frac{v - \sqrt{2}}{v + \sqrt{2}} | + C (∵ \int \frac{1}{x^{2} - a^{2}} d x = \frac{1}{2 a} \log | \frac{x - a}{x + a} | + C) \\ = \frac{1}{2 \cdot \sqrt{2}} \log | \frac{(t + \frac{1}{t} - \sqrt{2})}{(t + \frac{1}{t} + \sqrt{2})} | + C_{2} \end{aligned}

\begin{aligned} = \frac{1}{2 . \sqrt{2}} \log | \frac{(t^{2} + 1 - \sqrt{2} t)}{(\frac{t^{2} + 1 + \sqrt{2} t}{t})} | + C_{2} \\ = \frac{1}{2 . \sqrt{2}} \log | \frac{(t^{2} + 1 - \sqrt{2} t)}{(t^{2} + 1 + \sqrt{2} t)} | + C_{2} \dots \dots . . (I I I) \end{aligned}

Put the values of equ. (II) and (III) in equ.(I), We get,

\begin{aligned} I = \frac{1}{\sqrt{2}} \tan^{- 1} (\frac{(t^{2} - 1)}{(\sqrt{2} t)} + C_{1} - (\frac{1}{2 \cdot \sqrt{2}} \log | \frac{(t^{2} + 1 - \sqrt{2} t)}{(t^{2} + 1 + \sqrt{2} t)} | + C_{2}) \\ I = \frac{1}{\sqrt{2}} \tan^{- 1} (\frac{(t^{2} - 1)}{(\sqrt{2} t)} + C_{1} - \frac{1}{2 . \sqrt{2}} \log | \frac{(t^{2} + 1 - \sqrt{2} t)}{(t^{2} + 1 + \sqrt{2} t)} | - C_{2}) \\ I = \frac{1}{\sqrt{2}} \tan^{- 1} (\frac{(x - 1)}{(\sqrt{2} x)} - \frac{1}{2 . \sqrt{2}} \log | \frac{(x + 1 - \sqrt{2} x)}{(x + 1 + \sqrt{2} x)} | + C) (∵ t^{2} = x and c = c_{1} - c_{2}) Ans.. \end{aligned}

Indefinite Integrals Excercise 18.32 Question 7

Answer : -

\frac{1}{\sqrt{2}} \log | \frac{x + 2 - \sqrt{2 (x + 1)}}{x + 2 + \sqrt{2 (x + 1)}} | + C

Hint :- Use substitution and special integration formula.
Given :-

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

Sol : -

\begin{aligned} Let I = \int \frac{x}{(x^{2} + 2 x + 2) \sqrt{x + 1}} d x \\ Put, x + 1 = t^{2} \\ dx = 2 t dt then \end{aligned}

\begin{aligned} I = \int \frac{t^{2} - 1}{{{(t^{2} - 1)}^{2} + 2 (t^{2} - 1) + 2} \sqrt{t^{2}}} .2 tdt (∵ t^{2} - 1 = x) \\ = \int \frac{(t^{2} - 1)}{(t^{4} + 1 - 2 t^{2} + 2 t^{2} - 2 + 2) t} \cdot 2 t dt \\ = 2 \int \frac{t^{2} - 1}{(t^{4} + 1)} dt \end{aligned}

= 2 \int \frac{\frac{t^{2} - 1}{t^{2}}}{\frac{t^{4} + 1}{t^{2}}} dt

[Dividing num. And denom. By t2]

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

\begin{aligned} = 2 \int \frac{(1 - \frac{1}{t^{2}})}{t^{2} + \frac{1}{t^{2}} + 2 - 2} d t \\ = 2 \int \frac{(1 - \frac{1}{t^{2}})}{(t^{2} + \frac{1}{t^{2}} + 2 . t_{t}^{1}) - 2} d t \\ = 2 \int \frac{(1 - \frac{1}{t^{2}})}{{(t + \frac{1}{t})}^{2} - 2} dt [∵ (a + b)^{2} = a^{2} + 2 a b + b^{2}] \end{aligned}

\begin{aligned} Again, put t + \frac{1}{t} = u \\ \Rightarrow (1 - \frac{1}{t^{2}}) d t = du then \\ I = 2 \int \frac{d u}{u^{2} - (\sqrt{2})^{2}} \end{aligned}

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

\begin{aligned} = \frac{1}{\sqrt{2}} \log | \frac{t^{2} + 1 - \sqrt{2} t}{t^{2} + 1 + \sqrt{2} t} | + c \\ = \frac{1}{\cdot \sqrt{2}} \log | \frac{x + 1 + 1 - \sqrt{2} \cdot \sqrt{x + 1}}{x + 1 + 1 + \sqrt{2} \cdot \sqrt{x + 1}} | + c [∵ t = \sqrt{x + 1} or t^{2} = x + 1] \\ = \frac{1}{\sqrt{2}} \log | \frac{x + 2 - \cdot \sqrt{2 (x + 1)}}{x + 2 + \sqrt{2 (x + 1)}} | + c \end{aligned}

Indefinite Integrals Excercise 18.32 Question 8

Answer : -

- \frac{1}{\sqrt{2}} \log | (\frac{1}{x - 1} + \frac{1}{2}) + \sqrt{{(\frac{1}{x - 1} + \frac{1}{2})}^{2} + \frac{1}{4}} | + C

Hint :- Use substitution and special integration formula.
Given :-

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

Sol : - Let

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

Put,

x - 1 = \frac{1}{t}

\Rightarrow d x = - \frac{1}{t^{2}} d t

then,

\begin{aligned} \begin{aligned} I & = \int \frac{1}{\frac{1}{t} \sqrt{{(\frac{1}{t} + 1)}^{2} + 1}} (- \frac{1}{t^{2}}) d t \\ = - \int \frac{\frac{1}{t^{2}}}{\frac{1}{t} \sqrt{\frac{1}{t^{2}} + 1 + \frac{3}{t} + 1}} d t [∵ (a + b)^{2} = a^{2} + b^{2} + 2 a b] \end{aligned} \end{aligned}

\begin{aligned} = - \int \frac{\frac{1}{t^{2}}}{\frac{1}{t} \sqrt{\frac{1 + t^{2} + 2 t + t^{2}}{t^{2}}}} d t \\ = - \int \frac{\frac{1}{t^{2}}}{\frac{11}{t} \cdot \frac{1}{t} \sqrt{2 t^{2} + 2 t + 1}} d t \\ = - \int \frac{\frac{1}{t^{2}}}{\frac{1}{t^{2}} \sqrt{2 (t^{2} + t + \frac{1}{2})}} d t \end{aligned}

\begin{aligned} = - \int \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{t^{2} + 2 \cdot t \cdot \frac{1}{2} + {(\frac{1}{2})}^{2} - {(\frac{1}{2})}^{2} + \frac{1}{2}}} d t \\ = - \frac{1}{\sqrt{2}} \int \frac{1}{\sqrt{{(t + \frac{1}{2})}^{2} - \frac{1}{4} + \frac{1}{2}}} d t & [∵ a^{2} + b^{2} + 2 a b = (a + b)^{2}] \end{aligned}

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

Put,

t + \frac{1}{2} = u

\Rightarrow d t = d u

than

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

\begin{aligned} I = - \frac{1}{\sqrt{2}} \log | (t + \frac{1}{2}) + \sqrt{{(t + \frac{1}{2})}^{2} + \frac{1}{4}} | + c (∵ u = t + \frac{1}{2}) \\ I = - \frac{1}{\sqrt{2}} \log | (\frac{1}{x - 1} + \frac{1}{2}) + \sqrt{{(\frac{1}{x - 1} + \frac{1}{2})}^{2} + \frac{1}{4}} | + c (∵ (x - 1) = \frac{1}{t}) \end{aligned}

Indefinite Integrals Excercise 18.32 Question 9

Answer : -

- \log | (\frac{1}{x + 1} - \frac{1}{2}) + \sqrt{{(\frac{1}{x + 1} - \frac{1}{2})}^{2} + \frac{3}{4}} | + C

Hint :- Use substitution method and special integration formula to solve this integration
Given :-

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

Sol : - Let

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

Put,

x + 1 = \frac{1}{t}

\Rightarrow d x = - \frac{1}{t^{2}} d t

then,

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

\begin{aligned} = - \int \frac{\frac{1}{t^{2}}}{\frac{1}{t} \sqrt{\frac{1 - t + t^{2}}{t^{2}}}} d t \\ = - \int \frac{\frac{1}{t^{2}}}{{\frac{1}{t}}^{2} \bar{t} \sqrt{1 - t + t^{2}}} d t \\ = - \int \frac{\frac{1}{t^{2}}}{\frac{1}{t^{2}} \sqrt{t^{2} - t + 1}} d t \\ = - \int \frac{1}{\sqrt{t^{2} - 2 \cdot t_{2}^{\frac{1}{2}} + {(\frac{1}{2})}^{2} - {(\frac{1}{2})}^{2} + 1}} d t \end{aligned}

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

\begin{aligned} = - \int \frac{1}{\sqrt{{(t - \frac{1}{2})}^{2} - (\frac{- 3}{4})}} d t \\ = - \int \frac{1}{\sqrt{{(t - \frac{1}{2})}^{2} + (\frac{3}{4})}} d t \\ = - \int \frac{1}{\sqrt{{(t - \frac{1}{2})}^{2} + {(\frac{\sqrt{3}}{2})}^{2}}} d t \end{aligned}

Put,

t - \frac{1}{2} = u

\Rightarrow d t = d u

than,

\begin{aligned} I = - \int \frac{1}{\sqrt{u^{2} + {(\frac{\sqrt{3}}{2})}^{2}}} d u \\ = - \log | u + \sqrt{u^{2} + {(\frac{\sqrt{3}}{2})}^{2}} | + c [∵ \int \frac{1}{\sqrt{x^{2} + a^{2}}} d x = \log | x + \sqrt{x^{2} + a^{2}} | + c] \end{aligned}

\begin{aligned} = - \log | (t - \frac{1}{2}) + \sqrt{{(t - \frac{1}{2})}^{2} + \frac{3}{4}} | + c (∵ u = t - \frac{1}{2}) \\ = - \log | (\frac{1}{x + 1} - \frac{1}{2}) + \sqrt{{(\frac{1}{x + 1} - \frac{1}{2})}^{2} + \frac{3}{4}} | ∣ + c [∵ t = \frac{1}{x + 1}] \end{aligned}

Indefinite Integrals Excercise 18.32 Question 10

Answer : -

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

Hint :- Use substitution method and special integration formula.
Given :-

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

Sol : - Let

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

Put,

x = \frac{1}{t}

\Rightarrow d x = - \frac{1}{t^{2}} d t

then,

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

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

\begin{aligned} Put, 1 + t^{2} = u^{2} \\ \Rightarrow 2 t dt = 2 u du \\ I = - \int \frac{1}{{1 - (u^{2} - 1)] \sqrt{u^{2}}} \cdot u d u \\ I = - \int \frac{1}{(1 - u^{2} + 1) u} \cdot u du \\ I = - \int \frac{1}{2 - u^{2}} du \\ I = - \int \frac{1}{(\sqrt{2})^{2} - u^{2}} du \end{aligned}

\begin{array}{ll} I = - \frac{1}{2 \sqrt{2}} \log | \frac{\sqrt{2} + u}{\sqrt{2} - u} | + c & [∵ \int \frac{1}{a^{2} - x^{2}} d x = \frac{1}{2 a} \log | \frac{a + x}{a - x} | + c] \\ I = - \frac{1}{2 \sqrt{2}} \log | \frac{\sqrt{2} + \sqrt{1 + t^{2}}}{\sqrt{2} - \sqrt{1 + t^{2}}} | + c & [∵ 1 + t^{2} = u^{2} or u = \sqrt{1 + t^{2}}] \end{array}

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

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

Indefinite Integrals Excercise 18.32 Question 11

Answer : -

\frac{1}{\sqrt{3}} \tan^{- 1} (\sqrt{\frac{x^{2} + 1}{3}}) + C

Hint :- Use substitution method and special integration formula..
Given :-

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

Sol : - Let

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

\begin{array}{r} Put x^{2} + 1 = t^{2} \Rightarrow 2 xd x = 2 t dt \to x dx = t dt than, \\ I = \int \frac{1}{(t^{2} + 3) \sqrt{t^{2}}} t dt (∵ x^{2} + 1 = t^{2} \to x^{2} = t^{2} - 1 \\ \to x^{2} + 4 = t^{2} - 1 + 4 \to x^{2} + 4 = t^{2} + 3) \end{array}

\begin{aligned} I = \int \frac{1}{(t^{2} + 3) \cdot t} t dt \\ I = \int \frac{1}{t^{2} + 3} dt \\ I = \int \frac{1}{t^{2} + (\sqrt{3})^{2}} dt \end{aligned}

\begin{aligned} I = \frac{1}{\sqrt{3}} \tan^{- 1} (\frac{t}{\sqrt{2}}) + c [∵ \int \frac{1}{a^{2} + x^{2}} d x = \frac{1}{a} \tan^{- 1} (\frac{x}{a}) + c] \\ I = \frac{1}{\sqrt{3}} \tan^{- 1} (\frac{\sqrt{x^{2} + 1}}{\sqrt{3}}) + c \\ I = \frac{1}{\sqrt{3}} \tan^{- 1} (\sqrt{\frac{x^{2} + 1}{3}}) + c [∵ t = \sqrt{x^{2} + 1]} \end{aligned}

Indefinite Integrals Excercise 18.32 Question 12

Answer : -

- \frac{1}{\sqrt{2}} \tan^{- 1} (\sqrt{\frac{1 - x^{2}}{2 x^{2}}}) + C

Hint :- Use substitution method and special integration formula..
Given :-

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

Sol : - Let

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

Put

x = \frac{1}{t} \to d x = - \frac{1}{t^{2}} dt

then,

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

= - \int \frac{t}{(t^{2} + 1) \sqrt{t^{2} - 1}} dt

Put,

t^{2} - 1 = u^{2} \to 2 t dt = 2 u du \to t dt = u du

than,

\begin{aligned} I = - \int \frac{u d u}{(u^{2} + 1 + 1) \sqrt{u^{2}}} [∵ t^{2} - 1 = u^{2} \to t^{2} = u^{2} + 1] \\ = - \int \frac{u d u}{(u^{2} + 2) u} \\ = - \int \frac{d u}{u^{2} + (\sqrt{2})^{2}} \end{aligned}

\begin{aligned} = - \frac{1}{\sqrt{2}} \tan^{- 1} (\frac{u}{\sqrt{2}}) + c [∵ \int \frac{1}{a^{2} + x^{2}} d x = \frac{1}{a} \tan^{- 1} (\frac{x}{a}) + c] \\ = - \frac{1}{\sqrt{2}} \tan^{- 1} (\frac{\sqrt{t^{2} - 1}}{\sqrt{2}}) + c [∵ u^{2} = t^{2} - 1 \to u = \sqrt{t^{2} - 1}] \\ = - \frac{1}{\sqrt{2}} \tan^{- 1} (\sqrt{\frac{t^{2} - 1}{2}}) + c \end{aligned}

\begin{aligned} = - \frac{1}{\sqrt{2}} \tan^{- 1} (\sqrt{\frac{\frac{1}{x^{2}} - 1}{2}}) + c [∵ x = \frac{1}{t} \to t = \frac{1}{x}] \\ = - \frac{1}{\sqrt{2}} \tan^{- 1} (\sqrt{\frac{\frac{1 - x^{2}}{x^{2}}}{2}}) + c \\ = - \frac{1}{\sqrt{2}} \tan^{- 1} (\sqrt{\frac{1 - x^{2}}{2 x^{2}}}) + c \end{aligned}

Indefinite Integrals Excercise 18.32 Question 13

Answer : -

\frac{1}{2 \sqrt{33}} \log | \frac{\sqrt{11} x + \sqrt{3 x^{2} - 12}}{\sqrt{11} x - \sqrt{3 x^{2} - 12}} | + C

Hint :- Use substitution method and special integration formula.
Given :-

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

Sol : -

\begin{aligned} Let I = \int \frac{1}{(2 x^{2} + 3) \sqrt{x^{2} - 4}} d x \\ let, x = \frac{1}{t} \\ dx = - \frac{1}{t^{2}} dt then \end{aligned}

\begin{matrix} I = \int \frac{1}{(2 \cdot \frac{1}{t^{2}} + 3) \sqrt{\frac{1}{t^{2}} - 4}} (- \frac{1}{t^{2}} dt) \\ = - \int \frac{1}{(\frac{2 + 3 t^{2}}{t^{2}}) \sqrt{\frac{1 - 4 t^{2}}{t^{2}}}} \cdot \frac{1}{t^{2}} dt \\ = - \int \frac{\frac{1}{t^{2}}}{\frac{1}{t^{2}} (3 t^{2} + 2) \sqrt{1 - 4 t^{2}} \cdot \frac{1}{t}} dt \\ = - \int \frac{t}{(3 t^{2} + 2) \sqrt{1 - 4 t^{2}}} dt \end{matrix}

\begin{aligned} Put . 1 - 4 t^{2} = p^{2} \\ \Rightarrow - 4.2 t dt = 2 p dp \\ \Rightarrow tdt = - \frac{p d p}{4} \end{aligned}

I = - \int \frac{1}{{3 (\frac{1 - p^{2}}{4}) + 2} \cdot \sqrt{p^{2}}} \cdot - \frac{p d p}{4} [∵ 1 - 4 t^{2} = p^{2} \to 1 - p^{2} = 4 t^{2} \to t^{2} = \frac{1 - p^{2}}{4}]

\begin{aligned} I = \frac{1}{4} \int \frac{1}{\frac{3 - 3 p^{2} + 8}{4} \cdot p} \cdot pdp \\ I = \frac{1}{4} \int \frac{4}{11 - 3 p^{2}} dp \\ I = \frac{1}{3} \int \frac{1}{(\frac{11}{3} - p^{2})} dp \end{aligned}

\begin{aligned} I = \frac{1}{3} \int \frac{1}{{(\sqrt{\frac{11}{3}})}^{2} - p^{2}} dp \\ I = \frac{1}{3} \frac{1}{2 \cdot \frac{\sqrt{11}}{\sqrt{3}}} \log | \frac{\sqrt{\frac{11}{3}} + p}{\sqrt{\frac{11}{3}} - p} | + c [∵ \int \frac{1}{a^{2} - x^{2}} d x = \frac{1}{2 a} \log | \frac{a + x}{a - x} | \end{aligned}

\begin{aligned} = \frac{1}{2 \sqrt{3} \cdot \sqrt{11}} \log | \frac{\sqrt{\frac{11}{3}} + \sqrt{1 - 4 t^{2}}}{\sqrt{\frac{11}{3}} - \sqrt{1 - 4 t^{2}}} | + c [∵ p^{2} = 1 - 4 t^{2} \to p = \sqrt{1 - 4 t^{2}}] \\ = \frac{1}{2 \sqrt{33}} \log | \frac{\frac{\sqrt{11} + \sqrt{3} \sqrt{1 - 4 t^{2}}}{\sqrt{3}}}{\frac{\sqrt{11} - \sqrt{3} \sqrt{1 - 4 t^{2}}}{\sqrt{3}}} | + c \end{aligned}

\begin{aligned} = \frac{1}{2 \sqrt{33}} \log | \frac{\sqrt{11} + \sqrt{3 (1 - 4 t^{2})}}{\sqrt{11} - \sqrt{3 (1 - 4 t^{2})}} | + c \\ = \frac{1}{2 \sqrt{33}} \log | \frac{\sqrt{11} + \sqrt{3 (1 - 4 \frac{1}{x^{2}})}}{\sqrt{11} - \sqrt{3 (1 - 4 \frac{1}{x^{2}})}} | + c [∵ x = \frac{1}{t} \to t = \frac{1}{m}] \end{aligned}

\begin{aligned} = \frac{1}{2 \sqrt{33}} \log | \frac{\sqrt{11} + \sqrt{\frac{3 x^{2} - 12}{x^{2}}}}{\sqrt{11} - \sqrt{\frac{3 x^{2} - 12}{x^{2}}}} | + c \\ = \frac{1}{2 \sqrt{33}} \log | \frac{\sqrt{11} + \frac{1}{x} \sqrt{3 x^{2} - 12}}{\sqrt{11} - \frac{1}{x} \sqrt{3 x^{2} - 12}} | + c \end{aligned}

\begin{aligned} = \frac{1}{2 \sqrt{33}} \log | \frac{\frac{\sqrt{11} x + \sqrt{3 x^{2} - 12}}{x}}{\frac{\sqrt{\frac{11}{x} - \sqrt{3 x^{2} - 12}}}{x}} | + c \\ = \frac{1}{2 \sqrt{33}} \log | \frac{\sqrt{11} x + \sqrt{3 x^{2} - 12}}{\sqrt{11} x - \sqrt{3 x^{2} - 12}} | + c \end{aligned}

Indefinite Integrals Excercise 18.32 Question 14

Answer : -

\frac{1}{2 \sqrt{5}} \log | \frac{\sqrt{x^{2} + 9} - \sqrt{5}}{\sqrt{x^{2} + 9} + \sqrt{5}} | + C

Hint :- Use substitution method and special integration formula.
Given :-

\int \frac{x}{(x^{2} + 4) \sqrt{x^{2} + 9}} d x

Sol : -

\begin{aligned} Let I = \int \frac{x}{(x^{2} + 4) \sqrt{x^{2} + 9}} d x \\ put, x^{2} + 9 = u^{2} \to 2 x dx = 2 u du \\ \to x dx = u du than \\ I = \int \frac{u d u}{(u^{2} - 9 + 4) \sqrt{u^{2}}} \end{aligned}

\begin{aligned} I = \int \frac{u d u}{(u^{2} - 5) u} [∵ x^{2} + 9 = u^{2} \to x^{2} = u^{2} - 9] \\ I = \int \frac{u}{u^{2} - 5} d u \\ I = \int \frac{u}{u^{2} - (\sqrt{5})^{2}} d u \end{aligned}

\begin{aligned} I = \frac{1}{2 \cdot \sqrt{5}} \log | \frac{u - \sqrt{5}}{u + \sqrt{5}} | + c [∵ \int \frac{1}{x^{2} - a^{2}} dx = \frac{1}{2 a} \log | \frac{x - a}{x + a} | + c] \\ I = \frac{1}{2 \sqrt{5}} \log (\frac{\sqrt{x^{2} + 9} - \sqrt{5}}{\sqrt{x^{2} + 9} + \sqrt{5}}) + C [{\ddot{x}}^{2} + 9 = u^{2} \to u = \sqrt{x^{2} + 9}] \end{aligned}

Class 12 RD Sharma chapter 18 exercise 18.32 solution is prepared specifically for students who want to get good insight on the subject. It contains questions and answers in the same place making it easier for students to prepare and revise. Class 12 RD Sharma chapter 18 exercise 18.32 solution is a material that follows the CBSE syllabus and is updated to the latest edition.

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Some huge integrals close by theorems
A blend of sensible mathematical limits by using inadequate parts
Exactly when the denominator contains some reiterating straight factors, then the denominator contains unchangeable quadratic factors.
Blend of some interesting absurd arithmetical limits

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