Evaluate $\int_0^R e^{\frac{-r}{a}}r^2\mathrm dr$
Solution 1:
$$\int_{0}^{R}r^{2}e^{-\frac{r}{a}}dr=-aR^{2}e^{-\frac{R}{a}}+\int_{0}^{R}2are^{-\frac{r}{a}}dr=-aR^{2}e^{-\frac{R}{a}}-2a^{2}Re^{-\frac{R}{a}}+\int_{0}^{R}2a^{2}e^{-\frac{r}{a}}dr$$
$$=-aR^{2}e^{-\frac{R}{a}}-2a^{2}Re^{-\frac{R}{a}}+2a^{3}-2a^{3}e^{-\frac{R}{a}}$$
Integration by-parts
This technique is called Integration By parts. You cannot solve directly by substitution. You should also remember the ILATE RULE