Question

Evaluate: $\int \frac{1}{x(x^2+1)}dx$.

Answer 1

We are given,

$I=\int \frac{dx}{x(x^2+1)}$

Let $x=\tan \theta$

$dx={\sec }^2\theta d\theta$ (differentiation)

$dx=(1+{\tan }^2\theta )d\theta$

$\frac{dx}{(1+x^2)}=d\theta$

$I=\int \frac{d\theta }{\tan \theta }$

$=\int \cot \theta d\theta$

$I=\int \frac{\cos \theta }{\sin \theta }d\theta$

Let $\sin \theta =t$

$\cos d\theta =dt$ (differentiation)

$I=\int \frac{dt}{t}$

$I=In|t|+C$

$I=In|\sin \theta |+C$

Now, $\sin \theta =(\tan \theta ÷\sec \theta )$

$=x/\sqrt{1+x^2}$

$I=In\left(\frac{x}{\sqrt{1+x^2}}\right)+C$