# How do you evaluate the integral int x^2sqrt(x-3)?

Feb 11, 2017

I got:

$\frac{2}{35} {\left(x - 3\right)}^{\text{3/2}} \left(5 {x}^{2} + 12 x + 24\right) + C$

We don't really like radicals, so let's try letting $u = \sqrt{x - 3}$. Then:

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

Therefore, we have:

$\int 2 u {\left({u}^{2} + 3\right)}^{2} u \mathrm{du}$

$= 2 \int {u}^{2} {\left({u}^{2} + 3\right)}^{2} \mathrm{du}$

$= 2 \int {u}^{2} \left({u}^{4} + 6 {u}^{2} + 9\right) \mathrm{du}$

$= 2 \int {u}^{6} + 6 {u}^{4} + 9 {u}^{2} \mathrm{du}$

Now this is straightforward to integrate.

$= 2 \left({u}^{7} / 7 + \frac{6}{5} {u}^{5} + 3 {u}^{3}\right)$

$= \frac{2}{7} {u}^{7} + \frac{12}{5} {u}^{5} + 6 {u}^{3}$

Sub $u = \sqrt{x - 3}$ back in to get:

$= \frac{2}{7} {\left(x - 3\right)}^{\text{7/2" + 12/5(x-3)^"5/2" + 6(x-3)^"3/2}} + C$

This is acceptable, but we could also simplify further.

$= {\left(x - 3\right)}^{\text{3/2}} \left[\frac{2}{7} {\left(x - 3\right)}^{2} + \frac{12}{5} \left(x - 3\right) + 6\right] + C$

$= {\left(x - 3\right)}^{\text{3/2}} \left[\frac{2}{7} \left({x}^{2} - 6 x + 9\right) + \frac{12}{5} x - \frac{36}{5} + 6\right] + C$

$= {\left(x - 3\right)}^{\text{3/2}} \left(\frac{2}{7} {x}^{2} - \frac{12}{7} x + \frac{18}{7} + \frac{12}{5} x - \frac{36}{5} + \frac{30}{5}\right) + C$

$= {\left(x - 3\right)}^{\text{3/2}} \left(\frac{10}{35} {x}^{2} - \frac{60}{35} x + \frac{90}{35} + \frac{84}{75} x - \frac{252}{35} + \frac{210}{35}\right) + C$

$= {\left(x - 3\right)}^{\text{3/2}} \left(\frac{10}{35} {x}^{2} + \frac{24}{35} x + \frac{48}{35}\right) + C$

$= \textcolor{b l u e}{\frac{2}{35} {\left(x - 3\right)}^{\text{3/2}} \left(5 {x}^{2} + 12 x + 24\right) + C}$