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#### Need solution for RD Sharma Maths Class 12 Chapter 18 Indefinite Integrals Excercise 18.13 Question 4

Answer: $\frac{-1}{3} \frac{\left(1+x^{2}\right)^{\frac{3}{2}}}{x^{3}}+c \$

Hint: Use substitution method to solve this integral

Given: $\int \frac{\sqrt{1+x^{2}}}{x^{4}}dx$

Solution:

Let $I=\int \frac{\sqrt{1+x^{2}}}{x^{4}}dx$

Putting $x=\tan\theta$                                                   … (i)

$dx=\sec 2\theta d\theta$

Then

\begin{aligned} &I=\int \frac{\sqrt{1+\tan ^{2} \theta}}{\tan ^{4} \theta} \sec ^{2} \theta d \theta \\ &=\int \frac{\sqrt{\sec ^{2} \theta}}{\tan ^{4} \theta} \cdot \sec ^{2} \theta d \theta \quad\quad\quad\quad\quad\quad\quad\left[\begin{array}{l} \because \sec ^{2} \theta-\tan ^{2} \theta=1 \\ 1+\tan ^{2} \theta=\sec ^{2} \theta \end{array}\right] \\ &=\int \frac{\sec \theta \cdot \sec ^{2} \theta}{\tan ^{4} \theta} d \theta=\int \frac{\sec ^{3} \theta}{\tan ^{4} \theta} d \theta \\ &=\int \frac{1}{\cos ^{3} \theta \cdot \frac{\sin ^{4} \theta}{\cos ^{4} \theta}} d \theta=\int \frac{\cos ^{4} \theta}{\sin ^{4} \theta \cdot \cos ^{3} \theta} d \theta \\ &=\int \frac{\cos \theta}{\sin ^{4} \theta} d \theta \end{aligned}

Again putting    $\sin\theta=t$

\begin{aligned} &\Rightarrow \cos \theta d \theta=d t \text { then } \\ &I=\int \frac{1}{t^{4}} \cdot d t \\ &=\int t^{-4} d t=\frac{t^{-4+1}}{-4+1}+c \quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c\right] \\ &=\frac{t^{-3}}{-3}+c \\ &=\frac{-1}{3} t^{-3}+c \\ &=\frac{-1}{3 t^{3}}+c \quad \text { u. (ii) } \end{aligned}

Also from (i),  $x=\tan\theta$

\begin{aligned} &x^{2}=\tan ^{2} \theta\quad\quad\quad\quad \quad \text { [Squaring both sides] }\\ &\Rightarrow x^{2}=\frac{\sin ^{2} \theta}{\cos ^{2} \theta} \quad\quad\quad\quad\quad\quad\quad\quad\left[\tan \theta=\frac{\sin \theta}{\cos \theta}\right]\\ &\Rightarrow x^{2} \cos ^{2} \theta=\sin ^{2} \theta\\ &\Rightarrow x^{2}\left(1-\sin ^{2} \theta\right)=\sin ^{2} \theta \quad\quad\quad\quad\quad\left[\begin{array}{l} \sin ^{2} \theta+\cos ^{2} \theta=1 \\ \Rightarrow \sin ^{2} \theta=1-\cos ^{2} \theta \end{array}\right] \end{aligned}
\begin{aligned} &\Rightarrow x^{2}-x^{2} \sin ^{2} \theta=\sin ^{2} \theta\\ &\Rightarrow x^{2}=\sin ^{2} \theta+x^{2} \sin ^{2} \theta=\sin ^{2} \theta\left(1+x^{2}\right)\\ &\Rightarrow \sin ^{2} \theta=\frac{x^{2}}{1+x^{2}}\\ &\Rightarrow \sin \theta=\sqrt{\frac{x^{2}}{1+x^{2}}}\\ &\Rightarrow \sin \theta=\frac{x}{\sqrt{1+x^{2}}} \end{aligned}

Putting the value of sinθ from (iii) in (ii), we get

$I=\frac{-1}{3}\frac{1}{\left ( \sin\theta \right )^{3}}+c$

\begin{aligned} &=\frac{-1}{3} \frac{1}{\left(\frac{x}{\sqrt{1+x^{2}}}\right)^{3}}+c \\ &=\frac{-1}{3} \frac{1}{\frac{x^{3}}{\left(\sqrt{1+x^{2}}\right)^{3}}}+c \\ &=-\frac{1}{3} \frac{\left(1+x^{2}\right)^{\frac{3}{2}}}{x^{3}}+c \end{aligned}