Question

Prove that $\int \frac{x+3}{\sqrt{(x^2+2x+2)}}dx=\sqrt{x^2+2x+2}+2\log [x+1-{(x+1{)}^2+1}^{1/2}].$

Answer 1

Let $I=\int \frac{x+3}{\sqrt{x^2+2x+2}}dx=\int \frac{x+3}{\sqrt{{(x+1)}^2+1}}dx$
Put $x+1=t\Rightarrow dx=dt$
$I=\int \frac{t+2}{\sqrt{t^2+1}}dt$
Now put $t=\tan u$

$\Rightarrow dt={\sec }^{2}udu$
$\to I=\int (\tan u+2)\sec udu$

$=\int \tan u\sec udu+2\int \sec udu$

$=\sec u+2\log (\tan u+\sec u)$

$=\sqrt{1+{\tan }^{2}u}+2\log (\tan u+\sqrt{1+{\tan }^{2}u})$

$=\sqrt{t^2+1}+2\log (\sqrt{t^2+1}+t)$

$=\sqrt{{(x+1)}^2+1}+2\log (\sqrt{{(x+1)}^2}+x+1)$

$=\sqrt{x^2+2x+2}+2\log (\sqrt{x^2+2x+2}+x+1)$