# How do you integrate int5xsqrt(2x+3) using substitution?

Jul 17, 2016

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

#### Explanation:

This integral doesn't seem quiet obvious, so I will go through all the steps in high detail.

When evaluating integrals, it is sometimes a matter of trial and error, especially when it comes to u-substitutions.

For this particular integral, we may choose $u = 2 x + 3$ and see what happens.

If $u = 2 x + 3$, we'd have a $\mathrm{du}$ floating around somewhere - but there isn't. However, since there's a $5 x$ out front, we may be able to solve for $x$.

$u = 2 x + 3 \to \mathrm{du} = 2 \mathrm{dx} \to \frac{1}{2} \mathrm{du} = \mathrm{dx}$

Solving $u$ for $x$ we get

$u - 3 = 2 x \to \frac{u - 3}{2} = x$

We can, however, pull out a $\frac{1}{2}$ again, since it's just a constant.

Proceeding with the integration process, we can now rewrite our integral.

$\int 5 x \cdot \sqrt{2 x + 3} \mathrm{dx} = 5 \cdot \frac{1}{2} \cdot \frac{1}{2} \int \left(u - 3\right) {u}^{\frac{1}{2}} \mathrm{du}$

If we multiply everything into the parentheses and simplify we get

$\frac{5}{4} \int {u}^{\frac{3}{2}} - 3 {u}^{\frac{1}{2}} \mathrm{du} = \frac{5}{4} \left[\frac{2}{5} {u}^{\frac{5}{2}} - \frac{2}{\cancel{3}} \cdot \cancel{3} {u}^{\frac{3}{2}}\right] + C$

$= \frac{1}{2} {u}^{\frac{5}{2}} - \frac{5}{2} {u}^{\frac{3}{2}} + C$

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