Question

If$\int \frac{{\tan }^{-1}x}{x^4}dx=-\frac{{\tan }^{-1}x}{3x^3}\frac{1}{6x^2}-\frac{1}{3}\log |f\left(x\right)|+c$, then

• A. $f\left(x\right)=\frac{x}{\sqrt{1+x^2}}$
• B. $f\left(x\right)=\frac{-x}{\sqrt{1+x^2}}$
• C. $f\left(x\right)=\frac{2x}{\sqrt{1+x^2}}$
• D. $f\left(x\right)=-\frac{2x}{\sqrt{1+x^2}}$

Answer 1

Answer: (A)

$f\left(x\right)=\frac{x}{\sqrt{1+x^2}}$

Given, $\int \frac{{\tan }^{-1}x}{x^4}dx=-\frac{{\tan }^{-1}x}{3x^3}\frac{1}{6x^2}-\frac{1}{3}\log f\left(x\right)+c$ ....(1)
Now, $\int \frac{{\tan }^{-1}x}{x^4}dx=\int x^{-4}{\tan }^{-1}xdx$
Using Integration by parts, we get
$={\tan }^{-1}x\left(\frac{x^{-3}}{-3}\right)+\frac{1}{3}\int \frac{x^{-3}}{1+x^2}dx$
$=-\frac{1}{3x^3}{\tan }^{-1}x+\frac{1}{3}\int \frac{x^{-3}}{1+x^2}dx$
Now, substitute $x=\tan \theta \Rightarrow dx={\sec }^2\theta d\theta$
So, $=-\frac{1}{3x^3}{\tan }^{-1}x+\frac{1}{3}\int {\cot }^3\theta d\theta$
$=-\frac{1}{3x^3}{\tan }^{-1}x+\frac{1}{3}\int \cot \theta ({\text{cosec}}^2\theta -1)d\theta$
$=-\frac{1}{3x^3}{\tan }^{-1}x+\frac{1}{3}\int \cot \theta {\text{cosec}}^2\theta d\theta -\frac{1}{3}\int d\theta$
Substitute $\cot \theta =u\Rightarrow -{\text{cosec}}^2\theta d\theta =du$
$=-\frac{1}{3x^3}{\tan }^{-1}x-\frac{1}{3}\int udu-\frac{1}{3}\log |\sin \theta |+C$
$=-\frac{1}{3x^3}{\tan }^{-1}x-\frac{1}{3}\left[\frac{u^2}{2}\right]-\frac{1}{3}\log \left|\frac{x}{\sqrt{1+x^2}}\right|+C$
$=-\frac{1}{3x^3}{\tan }^{-1}x-\frac{1}{6{\tan }^2\theta }-\frac{1}{3}\log \left|\frac{x}{\sqrt{1+x^2}}\right|+C$
On comparing with given equation (1), we get
$f\left(x\right)=\frac{x}{\sqrt{1+x^2}}$