Question

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

Answer 1

Let, I $=\int \frac{x}{1+\sqrt{x}}dx$

Put $x=t^2\Rightarrow dx=2tdt$

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

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

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

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

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

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

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

$=\frac{2t^3}{3}-\frac{2t^2}{2}+2t-2\log |1+t|+c$

$=\frac{2t^3}{3}-t^2+2t-2\log |1+t|+c$

$=\frac{2{\left(\sqrt{x}\right)}^3}{3}-{\left(\sqrt{x}\right)}^2+2\sqrt{x}-2\log |1+\sqrt{x}|+c$, where $t=\sqrt{x}$

$=\frac{2x\sqrt{x}}{3}-x+2\sqrt{x}-2\log |1+\sqrt{x}|+c$