Question

$\int \frac{dx}{\sqrt{x^4+x^6}}$ is equal to

• A. $-\frac{\sqrt{1+x^2}}{x}+C$
• B. $\frac{\sqrt{1+x^2}}{x}+C$
• C. $-\frac{\sqrt{1-x^2}}{x}+C$
• D. $-\frac{\sqrt{x^2-1}}{x}+C$

Answer 1

The correct option is A $-\frac{\sqrt{1+x^2}}{x}+C$

Let $I=\int \frac{dx}{\sqrt{x^4+x^6}}=\int \frac{dx}{x^2\sqrt{x^2+1}}$
Put $x=\tan \theta$
$\Rightarrow dx={\sec }^2\theta d\theta$
$\therefore I=\int \frac{{\sec }^2\theta }{{\tan }^2\theta \sqrt{{\tan }^2\theta +1}}d\theta$
$=\int \frac{{\sec }^2\theta }{{\tan }^2\theta \cdot \sec \theta }d\theta$
$=\int \csc \theta \cot \theta d\theta$
$=-\csc \theta +C=\frac{-\sqrt{x^2+1}}{x}+C$