Evaluate $\int_0^R e^{\frac{-r}{a}}r^2\mathrm dr$

Question

Evaluate $\int_0^R e^{\frac{-r}{a}}r^2\mathrm dr$

$\beta=e^{-2r/a}$ and $d\beta=e^{2r/a}(-\frac{2}{a})\mathrm dr$ and $\ln \beta = -2r/a$

$$\int_0^R e^{\frac{-r}{a}}r^2\mathrm dr$$ $$\int_0^R\frac{a^2\ln \beta}{2}(-\frac{a}{2})\mathrm d\beta \tag{2}$$ $$\frac{-a^3}{4}\int_0^R \ln\beta\mathrm d\beta$$ $$\frac{-a^3}{4}[e^{-2R/a}\frac{-2R}{a}-e^{2R/a}]$$

But the answer is wrong if I can read solution sheet properly. I had taken all those variables properly but why the answer was wrong?

After seeing the comment, I tried to do it by my own. Latter I found expected answer. But I was wondering u-substitution wasn't wrong, if I looked properly, but why my answer didn't match? Is it necessary to do integral by parts when multiple variables are multiplying under integrand? equation (2) was very pretty easy to solve and u-substitution helped to bring it. I always look to solve things easily, it was also one of my method.

When you see integrals of the form $\int x^{n}e^{ax}dx$ you should try and apply Integration by parts. If you had an improper integral then you could have used Gamma function. — Mr. Gandalf Sauron, Dec 22 '21 at 11:12
@Mr.GandalfSauron Could you please give me a hint, which should I take as $u$? Cause the power of $r$ is disturbing me completely.. — Man, Dec 22 '21 at 11:14

Mr. Gandalf Sauron · Answer 1 · 2021-12-22T13:44:39.930

$$\int_{0}^{R}r^{2}e^{-\frac{r}{a}}dr=-aR^{2}e^{-\frac{R}{a}}+\int_{0}^{R}2are^{-\frac{r}{a}}dr=-aR^{2}e^{-\frac{R}{a}}-2a^{2}Re^{-\frac{R}{a}}+\int_{0}^{R}2a^{2}e^{-\frac{r}{a}}dr$$

$$=-aR^{2}e^{-\frac{R}{a}}-2a^{2}Re^{-\frac{R}{a}}+2a^{3}-2a^{3}e^{-\frac{R}{a}}$$

Integration by-parts

This technique is called Integration By parts. You cannot solve directly by substitution. You should also remember the ILATE RULE

Evaluate $\int_0^R e^{\frac{-r}{a}}r^2\mathrm dr$

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