# Integrating $\int_0^{2} x e^{\ln(2)x}$

How would I do this integral. I have tried the following.

$\int_0^{2} x e^{\ln(2)x}$

$u=x$ $dv=e^{\ln2(x)}$ $du=1$ $v=\frac{1}{\ln(2)}e^{\ln2(x)}$

$x\frac{1}{\ln(2)}e^{\ln2(x)}-\frac{1}{\ln(2)}\int e^{\ln2(x)}$

$x\frac{1}{\ln(2)}e^{\ln2(x)}-\frac{1}{ln(2)}\frac{1}{ln(2)}e^{\ln(2)(x)}$

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I see well what I meant is 2^x= e^ln(2)x –  Fernando Martinez Sep 6 '13 at 21:46
Dont just change $1$ title!!! –  Jean-Sébastien Sep 6 '13 at 21:52
Please, please, please don't forget to add the $\operatorname{d}\!x$ at the end of integrals when the variable is $x$. Leaving it off is a little like saying the area of a 2cm by 3cm rectangle is 2cm! The $\operatorname{d}\!x$ refers to the "width" of the areas. –  Fly by Night Sep 6 '13 at 22:00

$$F(x) = \dfrac x{\ln 2}e^{\ln(2) x} - \dfrac{e^{\ln(2) x}}{(\ln 2)^2} + C = -\ln 2\cdot e^{\ln(2) x}\left(x + \ln 2\right) + C$$ for example.
Now simply evaluate $F(2) - F(0)$