Question

Evaluate : $\int \frac{x}{\sqrt{x}+1}dx$

Answer 1

$\int \frac{x}{\sqrt{x}+1}dx$

Substituting $x=t^2$,

$dx=2tdt$

$=\int \frac{t^2}{t+1}\cdot 2tdt$

$=2\int \frac{t^3}{t+1}dt$

$=2\int \frac{t^3+1-1}{t+1}dt$

$=2\int [\frac{t^3+1}{t+1}-\frac{1}{t+1}]dt$

$=2\int [\frac{(t+1)(t^2+t+1)}{(t+1)}-\frac{1}{t+1}]dt$

$=2\int \left[\right(t^2+t+1)-\frac{1}{t+1}]dt$

$=2[\frac{t^3}{3}+\frac{t^2}{2}+t-\log \mid t+1\mid ]+c$

$=2[\frac{x\sqrt{x}}{3}+\frac{x}{2}+\sqrt{x}-\log \mid \sqrt{x}+1\mid ]+c$

$[\because x=t^2]$

Where $c$ is constant of integration
Hence, the required integration is
$2[\frac{x\sqrt{x}}{3}+\frac{x}{2}+\sqrt{x}-\log \mid \sqrt{x}+1\mid ]+c$