# How do you integrate int (2x^(3/2)+1)sqrtx using substitution?

Mar 29, 2018

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

#### Explanation:

Our goal with substitution is to find an expression whose differential appears in the integral, and represent it using the variable $u .$

We have $\int \left(2 {x}^{\frac{3}{2}} + 1\right) \sqrt{x} \mathrm{dx}$

In this case, $u = 2 {x}^{\frac{3}{2}} + 1$ is the best possible choice. Were we to choose $u = \sqrt{x} , \mathrm{du} = \frac{1}{2 \sqrt{x}} \mathrm{dx}$, and this differential is not even close to anything in the integral.

Calculate its differential:

$\mathrm{du} = \left(2\right) \left(\frac{3}{2}\right) {x}^{\frac{1}{2}} \mathrm{dx}$

$\mathrm{du} = 2 \sqrt{x} \mathrm{dx}$

Divide both sides by $2 :$

$\frac{1}{2} \mathrm{du} = \sqrt{x} \mathrm{dx}$

And we see $\sqrt{x} \mathrm{dx}$ does in fact show up in the integral. So, rewrite with the substitution:

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

Rewriting in terms of $x$ yields

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