# How do you find the integral int sqrt(4x-5)dx using substitution?

Feb 5, 2018

$\int \setminus \sqrt{4 x - 5} \setminus \mathrm{dx} = \frac{1}{6} \setminus {\left(4 x - 5\right)}^{\frac{3}{2}} + C$

#### Explanation:

We seek:

$I = \int \setminus \sqrt{4 x - 5} \setminus \mathrm{dx}$

We can perform a substitution:

$u = 4 x - 5 \implies \frac{\mathrm{du}}{\mathrm{dx}} = 4$

Then, we can substitute into the integral, as follows:

$I = \frac{1}{4} \setminus \int \setminus \sqrt{4 x - 5} \setminus \left(4\right) \setminus \mathrm{dx}$
$\setminus \setminus = \frac{1}{4} \setminus \int \setminus \sqrt{u} \setminus \mathrm{du}$
$\setminus \setminus = \frac{1}{4} \setminus \int \setminus {u}^{\frac{1}{2}} \setminus \mathrm{du}$

Which is a standard integral, so using the power rule:

$I = \frac{1}{4} \setminus {u}^{\frac{3}{2}} / \left(\frac{3}{2}\right) + C$
$\setminus \setminus = \frac{1}{4} \setminus \frac{2}{3} \setminus {u}^{\frac{3}{2}} + C$
$\setminus \setminus = \frac{1}{6} \setminus {u}^{\frac{3}{2}} + C$

And then restoring the substitution, we find that:

$I = \frac{1}{6} \setminus {\left(4 x - 5\right)}^{\frac{3}{2}} + C$