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Can you guys help me integrate $ \int xe^{-2x} dx$ using integration by parts?

So far I got an answer using this $$u = x \qquad dv = e^{-2x}dx \\ du = dx \qquad v = \frac{-e^{-2x}}{2} $$ so that would mean that $$ - \frac{-xe^{-2x}}{2} - \int \frac{-e^{-2x}}{2}dx$$ and the final answer would have been $$ \frac{-xe^{-2x}}{2} - e^{-2x} + c$$ is this correct or should i have interchanged my $u$ and $dv$?

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  • $\begingroup$ Can you do the integral which results from the IBP formula? If you can then the working was successful, if not then one of the things you could try is to interchange $u$ and $v$. $\endgroup$
    – David
    Jan 7, 2015 at 9:37
  • $\begingroup$ see this link, en.wikipedia.org/wiki/Integration_by_parts just choose now the right functions for $u(x)$ and $v(x)$ Hint for this: you see in the integral that there appears $u'(x)$, so it would be nice if this would be 1 $\endgroup$
    – Bobby
    Jan 7, 2015 at 9:38
  • $\begingroup$ yes i could and i got an answer but i tried interchanging the two and i got a different answer, the problem would then be to decide which one is correct $\endgroup$
    – sheila
    Jan 7, 2015 at 9:38
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    $\begingroup$ Please check your work for $\int -\frac{e^{-2x}}{2}$ $\endgroup$
    – paw88789
    Jan 7, 2015 at 9:39
  • $\begingroup$ should it be [-e^(-2x)]/4??? sorry i'm very confused right now $\endgroup$
    – sheila
    Jan 7, 2015 at 9:42

2 Answers 2

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You begin parts from the exponential, that is: $$\begin{aligned} \int x e^{-2x}\,dx &=-\frac{1}{2} xe^{-2x}+\frac{1}{2}\int (x)' e^{-2x}\,dx \\ &=-\frac{1}{2}xe^{-2x}+\frac{1}{2}\int e^{-2x}\,dx \\ &= -\frac{1}{2}xe^{-2x}-\frac{1}{4}e^{-2x}+c, \;\; c \in \mathbb{R}\\ \end{aligned}$$

done.

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$$\left(-\frac{xe^{-2x}}{2} - e^{-2x} + c\right)'=-\frac{e^{-2x}}2+xe^{-2x}+2e^{-2x}.$$

You visibly messed up with a factor...

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