A few definite integrals a) $\displaystyle \int_{-1}^0 x \sqrt{2 - 5x}dx. $
I think I should use some change of variables here, but it didn't work the way I tried to use it. 
b) $\displaystyle \int_{0}^{2} \frac{1}{(x + 2)(x + 3)}dx.$

Another way is by a rationalizing substitution:
$$
\begin{align}
u & = \sqrt{2-5x} \\[8pt]
u^2 & = 2 - 5x \\[8pt]
2u\,du & = -5\,dx \\[8pt]
\frac{-2}{5} u \, du & = dx \\[8pt]
x & = \frac{2-u^2}{5}
\end{align}
$$
When $x=-1$ then $u=\sqrt{7}$ and when $x=0$ then $u=\sqrt{2}$.  So
$$
\int_{-1}^0 x \sqrt{2-5x} \, dx = \int_\sqrt{7}^\sqrt{2} \frac{2-u^2}{5}\cdot u\left(\frac{-2}{5} u\, du \right) = \frac{4}{25}\int_\sqrt{2}^\sqrt{7}(2u^2-u^4)\,du
$$
