Integral of $x^2\sqrt{5+x}\ dx$

Question

I have the following integral to solve, with my working out below. This is a bit more complicated than I am used to, so I'm hoping for some feedback as I'm not sure if my process & solution are correct:

$\int x^2\sqrt{5+x}\ dx$

Let $u = 5+x$

$du = dx$

$x = u-5$

$=\int (u-5)^2\sqrt{u}\ du$

$=\int (u^2-10u+25)\sqrt{u}\ du$

$=\int u-10u^{3/2}+25u^{1/2}\ du$

$= \frac{u^2}{2} - 4u^{5/2} + \frac{50}{3}u^{3/2}\ +C$

$= \frac{(5+x)^2}{2} - 4(5+x)^{5/2} + \frac{50}{3}(5+x)^{3/2}\ +C$

Answer 1

Note that $$\int(u^2-10u+25)\sqrt u\ du=\int(u^{\color{red}{\frac 52}}-10u^{\frac 32}+25u^{\frac 12})du$$ because $u^2\times \sqrt u=u^2\times u^{\frac 12}=u^{2\color{red}{+}\frac 12}$ (not $u^{2\times \frac 12}$)

Answer 2

Replace $ x+5 $ by $ u^2$ , this will turn the integral to much simpler format $$ dx = 2u du $$ $$ \int (u^2 -5)^2 \times 2u^2 du$$