Ughh confusing integrals ...
1) Integrate $\displaystyle x^2sqrt(x+2) dx$
2) Integrate $\displaystyle x/(2-x^2) dx$
Observe that $\displaystyle d(x^2) = 2x \, dx$
So
$\displaystyle \int \frac{x}{2-x^2}\, dx = -\frac12 \int \frac{-2x}{2-x^2}\, dx = -\frac12 \int \frac1{2-x^2}\, d(2 - x^2)$
If this method looks weird to you, use the substitution u = 2 - x^2. I have done the same thing, but avoided introducing another variable
(2) Let $\displaystyle {\color{blue}u = 2 - x^{2}}$ whose derivative is $\displaystyle du = -2x dx \: \Rightarrow \: {\color{red}x dx = -\frac{du}{2}}$
So the integral: $\displaystyle \int \frac{{\color{red}xdx}}{{\color{blue}2-x^{2}}} = \int \frac{-\frac{du}{2}}{u} = -\frac{1}{2} \int \frac{du}{u}$