Computing $\int_0^1 xe^{2x}\, dx$ 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?
 A: 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.
A: 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:
$$
\int_0^1 x\mathrm{e}^{2x}\,dx=\left[{1\over 2}x\mathrm{e}^{2x}-{1\over 4}\mathrm{e}^{2x}\right]_0^1={1\over 4}\mathrm{e}^2+{1\over 4}.
$$
