How do you use integration by parts to find $\int x e^{x} d x$ $intxe^x dx$ ?

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How do you use integration by parts to find $\int x e^{x} d x$ $intxe^x dx$ ?

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Answer 1 · 2014-10-07T11:17:03

To integrate by parts, we have to pick a $u$ $u$ and $d v$ $dv$ such that

$\int u d v = u v - \int v d u$ $int u dv= uv-int v du$

We can pick what $u$ $u$ and $d v$ $dv$ are, though we should typically try to pick $u$ $u$ such that $d u$ $du$ is "simpler". (By the way, $d u$ $du$ just means the derivative of $u$ $u$ , and $d v$ $dv$ just means derivative of $v$ $v$ )

So, from $\int u d v$ $int u dv$ , let's pick $u = x$ $u=x$ and $d v = e^{x}$ $dv=e^x$ . It is helpful to fill in all of the parts you will use throughout the problem before you start integrating:

$u = x$ $u=x$
$\frac{d u}{d x} = 1$ $(du)/dx=1$ so $d u = 1 d x = d x$ $du=1dx=dx$

$d v = e^{x}$ $dv=e^x$
$v = \int d v = \int e^{x} d x = e^{x}$ $v=int dv= int e^x dx =e^x$

Now let's plug everything back into the original formula.

$\int u d v = u v - \int v d u$ $int u dv= uv-int v du$
$\int x e^{x} d x = x e^{x} - \int e^{x} d x$ $int xe^x dx= xe^x-int e^xdx$
You can integrate the last part quite simply now, to get:

$\int x e^{x} d x = x e^{x} - \int e^{x} d x = x e^{x} - e^{x} + c$ $int xe^x dx= xe^x-int e^xdx= xe^x-e^x +c$

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How do you use integration by parts to find $\int x e^{x} d x$ $intxe^x dx$ ?

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