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My approach to compute $\int_0^1 xe^{2x}\, dx$ is via integration by parts by setting $$ u'(x)=x^2,\qquad v(x)=e^{2x} $$ which gives me $$ \int_0^1xe^{2x}\, dx=\frac{e^2}{2}-\int_0^1x^2e^{2x}\, dx. $$ Then, doinf again integration by parts by setting $$ u'(x)=x^2,\qquad v(x)=e^{2x}, $$ I get $$ \int_0^1 x^2e^{2x}\, dx=\frac{e^3}{3}-\frac{2}{3}\int_0^1x^3e^{2x}\, dx $$ which gives me $$ \int_0^1 xe^{2x}\, dx=\frac{e^2}{6}+\frac{2}{3}\int_0^1 x^3e^{2x}\, dx. $$

I guess now I have to do another integration by parts and so on but this won't come to an end. So what am I doing wrong?

John_Doe
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    Try integrating by parts in the other direction. You want the polynomial term to vanish, hence you should differentiate that. – Daniel Fischer Nov 02 '16 at 13:02
  • Ah, okay. Thanks! Then it is an easy computation! $$\int_0^1 xe^{2x}, dx=\frac{e^2}{4}+\frac{1}{4}$$ – John_Doe Nov 02 '16 at 13:06
  • When you try integration by parts, and the problem gets worse instead of better, that may mean you should try integration by parts in the other direction. As you saw, that is the case here. Of course, some integrals get worse in both directions, so integration by parts does not help. – GEdgar Nov 02 '16 at 13:16

2 Answers2

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There is a rule of thumb, known as the ILATE or the LIATE rule, where

I - Inverse trigonometric, L - Logarithmic, A - Algerbraic, T - Trigonometric, E - Exponential,

It suggests that the type of function that appears first in the acronym be taken as $v(x)$ and the other as $u'(x)$.

In this case, $x$ is algebraic and $e^{2x}$ is exponential. This suggests a reverse order from what you chose.

$$I=\int_0^1xe^{2x}dx=\frac{e^2}{2}-\frac12\int_0^1e^{2x}dx$$

which can be easily simplified.

Also see this question and this question.

Community
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GoodDeeds
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You can easily compute the integral by parts: recall that $\int fg'=fg-\int f'g$. Then you choose $f(x)=x$, so that $f'(x)=1$, and $g'(x)=\mathrm{e}^{2x}$, getting thus $g(x)={1\over 2}\mathrm{e}^{2x}$. Compute $$ \int x\mathrm{e}^{2x}\,dx={1\over 2}x\mathrm{e}^{2x}-{1\over 2}\int \mathrm{e}^{2x}\,dx={1\over 2}x\mathrm{e}^{2x}-{1\over 4}\mathrm{e}^{2x}+C. $$ Thus you have: $$ \int_0^1 x\mathrm{e}^{2x}\,dx=\left[{1\over 2}x\mathrm{e}^{2x}-{1\over 4}\mathrm{e}^{2x}\right]_0^1={1\over 4}\mathrm{e}^2+{1\over 4}. $$

EM90
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