Question

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

• A. $-\frac{1}{3}\frac{(x^2+1{)}^{3/2}}{x^3}+C$
• B. $x^3(x^2+1{)}^{-1/2}+C$
• C. $\frac{\sqrt{x^2+1}}{x^2}+C$
• D. $-\frac{1}{3}\frac{(x^2+1{)}^{3/2}}{x^2}+C$

Answer 1

The correct option is A $-\frac{1}{3}\frac{(x^2+1{)}^{3/2}}{x^3}+C$

$I=\int \frac{\sqrt{x^2+1}}{x^4}dx$
Put $x=\tan \theta$
$\sqrt{x^2+1}=\sec \theta$
$dx={\sec }^2\theta d\theta$
$\therefore I=\int \frac{\sec \theta {\sec }^2\theta }{{\tan }^4\theta }d\theta$
$=\int \frac{\cos \theta }{{\sin }^4\theta }d\theta$
Put $\sin \theta =t$
$\cos \theta d\theta =dt$
$I=\int \frac{1}{t^4}dt$
$=-\frac{1}{3}\frac{1}{{\sin }^3\theta }+C$
$=-\frac{1}{3}\frac{{\sec }^3\theta }{{\tan }^3\theta }+C$
$=-\frac{1}{3}\frac{(x^2+1{)}^{3/2}}{x^3}+C$