Question

Evaluate :

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

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

Answer 1

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

Let $I=\int \frac{dx}{x\sqrt{1+x^2}}$
Substitute $x=\tan u\Rightarrow dx={\sec }^{2}udu$
$\therefore I=\int \cot u\csc udu=\int \frac{\cos u}{{\sin }^{2}u}du$
Substitute $s=\sin u\Rightarrow ds=\cos udu$
$\therefore I=\int \frac{1}{s^2}ds=-\frac{1}{s}+c$
$=-\frac{\sqrt{1+x^2}}{x}+c=-\sqrt{1+\frac{1}{x^2}}+c$