Question

If $I=\int \frac{dx}{x^4\sqrt{1+x^2}}$, then I equals

• A. $\frac{\sqrt{x^2+1}}{x}-\frac{1}{2x^2}+C$
• B. $\frac{\sqrt{1+x^2}}{x}-\frac{1}{2x^3}+C$
• C. $-\frac{\sqrt{1+x^2}}{x}+\frac{2x}{\sqrt{1+x^2}}+C$
• D. none of these

Answer 1

The correct option is D none of these

$I=\int \frac{1}{x^4\sqrt{1+x^2}}dx$
Put $x=\tan t\Rightarrow dx={\sec }^{2}tdt$
$\therefore I=\int {\cot }^{3}t\csc tdx=\int cott\csc t\left({\csc }^{2}t-1\right)dt$
Put $u=\csc t\Rightarrow du=-cot\csc tdu$
$\therefore I=-\int (u^2-1)du=-\frac{u^3}{3}+u+c$
$=\csc \left({\tan }^{-1}x\right)-\frac{1}{3}\csc {\left({\tan }^{-1}x\right)}^3+c$
$=\frac{\sqrt{x^2+1}(2x^2-1)}{3x^3}+c$