Question #24569

Aug 7, 2016

$\frac{2 \sqrt{x + a} \left(x - 2 a\right)}{3} + C$

Explanation:

We have:

$\int \frac{x}{\sqrt{x + a}} \mathrm{dx}$

If we want to use substitution, a good bet would be to let $u = x + a$. This also would mean that $\mathrm{du} = \mathrm{dx}$. The only issue left is that we've accounted for both $\mathrm{dx}$ and $\sqrt{x + a}$ in the denominator, and still have the $x$ in the numerator left over.

To deal with this, use the original substitution $u = x + a$ to write $x$ as $x = u - a$. Then, we have:

$\int \frac{x}{\sqrt{x + a}} \mathrm{dx} = \int \frac{u - a}{\sqrt{u}} \mathrm{du}$

At this point, split the integral through subtraction:

$\int \frac{u - a}{\sqrt{u}} \mathrm{du} = \int \frac{u}{\sqrt{u}} \mathrm{du} - \int \frac{a}{\sqrt{u}} \mathrm{du}$

Rewrite using powers:

$\int \frac{u}{\sqrt{u}} \mathrm{du} - \int \frac{a}{\sqrt{u}} \mathrm{du} = \int {u}^{\frac{1}{2}} \mathrm{du} - a \int {u}^{- \frac{1}{2}} \mathrm{du}$

Integrate both using the power rule for integrals:

$\int {u}^{\frac{1}{2}} \mathrm{du} - a \int {u}^{- \frac{1}{2}} \mathrm{du} = {u}^{\frac{3}{2}} / \left(\frac{3}{2}\right) - a \left({u}^{\frac{1}{2}} / \left(\frac{1}{2}\right)\right) + C$

$= \frac{2 {u}^{\frac{3}{2}}}{3} - 2 a {u}^{\frac{1}{2}} + C$

$= \frac{2 \sqrt{u} \left(u - 3 a\right)}{3} + C$

Since $u = x + a$:

$= \frac{2 \sqrt{x + a} \left(x + a - 3 a\right)}{3} + C$

$= \frac{2 \sqrt{x + a} \left(x - 2 a\right)}{3} + C$