Question #0f959

Sep 18, 2016

$\frac{4}{3} {\left({x}^{2} + x + 1\right)}^{\frac{3}{2}} + C$

Explanation:

We have:

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

You may not realize it yet, but this is set up quite well for substitution. We will use the substitution $u = {x}^{2} + x + 1$. Note that this implies that $\mathrm{du} = \left(2 x + 1\right) \mathrm{dx}$.

Factor a $2$ from the $\left(4 x + 2\right)$ term:

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

Notice both our $u$ and $\mathrm{du}$ terms are present:

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

$= 2 \int {u}^{\frac{1}{2}} \mathrm{du}$

Using the rule $\int {u}^{n} \mathrm{du} = {u}^{n + 1} / \left(n + 1\right) + C$:

$= 2 \left({u}^{\frac{3}{2}} / \left(\frac{3}{2}\right)\right) + C$

$= \frac{4}{3} {u}^{\frac{3}{2}} + C$

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