# How do you find the integral for (x)/(sqrt(1+2x))dx?

Apr 1, 2015

I would probably use parts, but since this is asked in Integration by substitution, I'll do it that way.

$\int \frac{x}{\sqrt{1 + 2 x}} \mathrm{dx}$

Let $u = 1 + 2 x$. This makes $\mathrm{du} = 2 \mathrm{dx}$ and $\mathrm{dx} = \frac{1}{2} \mathrm{du}$.

The substitution is not complete until we also note that
$u = 1 + 2 x$ makes $x = \frac{u - 1}{2}$

$\int \frac{x}{\sqrt{1 + 2 x}} \mathrm{dx} = \int x \frac{1}{\sqrt{1 + 2 x}} \cdot \mathrm{dx}$

$= \int \frac{u - 1}{2} \cdot \frac{1}{\sqrt{u}} \cdot \frac{1}{2} \mathrm{du} = \frac{1}{4} \int \left(u - 1\right) {u}^{- \frac{1}{2}} \mathrm{du}$

$= \frac{1}{4} \int \left({u}^{\frac{1}{2}} - {u}^{- \frac{1}{2}}\right) \mathrm{du}$

You can probably finish this yourself.

$= \frac{1}{4} \left[\frac{2}{3} {u}^{\frac{3}{2}} - \frac{2}{1} {u}^{\frac{1}{2}}\right] + C = \frac{1}{6} \left[{u}^{\frac{3}{2}} - 3 {u}^{\frac{1}{2}}\right] + C$

$\int \frac{x}{\sqrt{1 + 2 x}} \mathrm{dx} = \frac{1}{6} \left[{\sqrt{1 + 2 x}}^{3} - 3 \sqrt{1 + 2 x}\right] + C$