Question

Solve:

$\int \frac{1}{\left(x\sqrt{x^2+a^2}\right)}dx$

Answer 1

We have,

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

Let

$t^2=x^2+a^2$

$2t.dt=2x.dx$

$t.dt=x.dx$

Therefore,

$I=\int \frac{t}{\left(t^2-a^2\right)t}dt$

$I=\int \frac{1}{\left(t^2-a^2\right)}dt$

$I=\frac{1}{2a}\log \left[\frac{t-a}{t+a}\right]+C$

On putting the value of $t$, we get

$I=\frac{1}{2a}\log \left[\frac{\sqrt{x^2+a^2}-a}{\sqrt{x^2+a^2}+a}\right]+C$

Hence, this is the answer.