# How do you evaluate the integral int x/sqrt(4x+1)?

Apr 10, 2017

$\int \frac{x}{\sqrt{4 x + 1}} \mathrm{dx} = \frac{2 x - 1}{12} \sqrt{4 x + 1} + C$

#### Explanation:

Note that:

$\frac{1}{\sqrt{4 x + 1}} = \frac{1}{2} \frac{d}{\mathrm{dx}} \sqrt{4 x + 1}$

So we can write the integral as:

$\int \frac{x}{\sqrt{4 x + 1}} \mathrm{dx} = \frac{1}{2} \int x d \left(\sqrt{4 x + 1}\right)$

and integrate by parts:

$\int \frac{x}{\sqrt{4 x + 1}} \mathrm{dx} = \frac{1}{2} x \sqrt{4 x + 1} - \frac{1}{2} \int \sqrt{4 x + 1} \mathrm{dx}$

The resulting integral can be resolved directly using the power rule:

$\int \frac{x}{\sqrt{4 x + 1}} \mathrm{dx} = \frac{1}{2} x \sqrt{4 x + 1} - \frac{1}{8} \int {\left(4 x + 1\right)}^{\frac{1}{2}} d \left(4 x + 1\right)$

$\int \frac{x}{\sqrt{4 x + 1}} \mathrm{dx} = \frac{1}{2} x \sqrt{4 x + 1} - \frac{1}{8} {\left(4 x + 1\right)}^{\frac{3}{2}} / \left(\frac{3}{2}\right) + C$

and simplifying:

$\int \frac{x}{\sqrt{4 x + 1}} \mathrm{dx} = \frac{1}{2} x \sqrt{4 x + 1} - \frac{1}{12} \left(4 x + 1\right) \sqrt{4 x + 1} + C$

$\int \frac{x}{\sqrt{4 x + 1}} \mathrm{dx} = \left(\frac{1}{2} x - \frac{1}{3} x - \frac{1}{12}\right) \sqrt{4 x + 1} + C$

$\int \frac{x}{\sqrt{4 x + 1}} \mathrm{dx} = \frac{2 x - 1}{12} \sqrt{4 x + 1} + C$