Chapter 5.7, Problem 13E

To determine

To evaluate the integral ${\displaystyle \int \frac{1}{x^2\sqrt{x^2+4}}}dx$

Answer to Problem 13E

The value is $-\frac{\sqrt{x^2+4}}{4x}+C$

Explanation of Solution

Given information :

The integral is ${\displaystyle \int \frac{1}{x^2\sqrt{x^2+4}}}dx$

Formula used :

  $\frac{d}{d\theta }\left(\tan \theta \right)={\sec }^2\theta$ , $\frac{d}{d\theta }\left(\sin \theta \right)=\cos \theta$ , ${\displaystyle \int x^n=\frac{x^n}{n+1}+C}$

Calculation :

The integral is ${\displaystyle \int \frac{1}{x^2\sqrt{x^2+4}}}dx$

Given $x=2\tan \theta$ , therefore $\sin \theta =\frac{x}{\sqrt{x^2+4}}$

Differentiating both side , $dx=2{\sec }^2\theta d\theta$

Substituting $x=2\tan \theta$ and $dx=2{\sec }^2\theta d\theta$ in given integral

  ${\displaystyle \int \frac{1}{4{\tan }^2\theta \sqrt{4{\tan }^2\theta +4}}}\times 2{\sec }^2\theta d\theta ={\displaystyle \int \frac{1}{4{\tan }^2\theta \times 2\sqrt{{\tan }^2\theta +1}}}\times 2{\sec }^2\theta d\theta =\frac{1}{4}{\displaystyle \int \frac{1}{{\tan }^2\theta \sec \theta }}{\sec }^2\theta d\theta =\frac{1}{4}{\displaystyle \int \frac{\sec \theta }{{\tan }^2\theta }}d\theta =\frac{1}{4}{\displaystyle \int \frac{\cos \theta }{{\sin }^2\theta }}d\theta$

Using Substitution Rule , Let $\sin \theta =t$

Differentiating both side , $\cos \theta d\theta =dt$

  $=\frac{1}{4}{\displaystyle \int \frac{1}{t^2}}dt=\frac{1}{4}\left(\frac{-1}{t}\right)+C=-\frac{\sqrt{x^2+4}}{4x}+C$ , Where $C$ is constant of integration

Hence the value of ${\displaystyle \int \frac{1}{x^2\sqrt{x^2+4}}}dx$ is $-\frac{\sqrt{x^2+4}}{4x}+C$