Question

Integrate with respect to $x$:
$\frac{x}{\sqrt{x+2}}$

Answer 1

Consider the given integral.

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

Put,

$u=x+2\Rightarrow du=dx$

Therefore,

$I=\int \frac{u-2}{\sqrt{u}}du$

$I=\int (\sqrt{u}-\frac{2}{\sqrt{u}})du$

$I=\int \sqrt{u}du-2\int \frac{1}{\sqrt{u}}du$

$I=\frac{2}{3}{\left(u\right)}^{3/2}-4\sqrt{u}+C$

Now, return the value of $x$.

$I=\frac{2}{3}{(x+2)}^{3/2}-4\sqrt{x+2}+C$

$I=\frac{2\sqrt{x+2}(x+2-6)}{3}+C$

$I=\frac{2\sqrt{x+2}(x-4)}{3}+C$

Hence, this is the required value of the integral.