Question

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

Answer 1

We have,

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

Let
$t=x^2+4x+1$

$\frac{dt}{dx}=2x+4$

$\frac{dt}{2}=(x+2)dx$

Therefore,

$I=\frac{1}{2}\int \frac{1}{\sqrt{t}}dt$

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

$I=\sqrt{t}+C$

On putting the value of $t$, we get

$I=\sqrt{x^2+4x+1}+C$

Hence, this is the answer.