Question

What is $\int \frac{dx}{\sqrt{4+x^2}}$ equal to?

• A. $ln\left|\frac{\sqrt{4+x^2}+x}{2}\right|+c$
• B. $ln\left|\frac{\sqrt{4+x^2}-x}{2}\right|+c$
• C. ${\sin }^{-1}\left(\frac{x}{2}\right)+c$
• D. None of the above
  where $c$ is an arbitrary constant

Answer 1

The correct option is A $ln\left|\frac{\sqrt{4+x^2}+x}{2}\right|+c$

$Q\int \frac{1}{\sqrt{x}+4}dx$

put $x=2\tan \theta .$

$dx=2{\sec }^2\theta d\theta$

$\Rightarrow \int \frac{1}{\sqrt{4+x^2}}dx=\int \frac{1}{\sqrt{4+4{\tan }^2\theta }}2{\sec }^2\theta d\theta$

$=\int \frac{{\sec }^2\theta }{\sec \theta }d\theta =\int \sec \theta d\theta =ln|\sec \theta +\tan \theta |+c.$

As $x=2\tan \theta$

$\Rightarrow \frac{x}{2}=\tan \theta \Rightarrow \frac{x}{2}=\sqrt{se{c}^2\theta 1}$

or $\frac{x^2}{4}+1={\sec }^2\theta$ or $\sqrt{x^2}4+1=\sec \theta$

$\Rightarrow ln|\sec \theta +\tan \theta |+c$

$=ln|\sqrt{\frac{x^2}{4}+1}+\frac{x}{2}|+c$

$=ln\left|\frac{\sqrt{x^2+4}+x}{2}\right|+c$

$=ln\left|\frac{\sqrt{4+x^2}+x}{2}\right|+c$