What is the antiderivative of $x e^{x}$ $xe^x$ ?

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Answer 1 · 2016-07-08T07:39:11

Let $u = x$ $u=x$ and $d v = e^{x} d x$ $dv= e^(x) dx$ , so $d u = d x$ $du=dx$ and $v = e^{x}$ $v=e^(x)$

Knowing that

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

You can see that

$\int x e^{x} d x$ $int x e^(x) dx$
$= x e^{x} - \int e^{x} d x$ $= xe^(x)-int e^x dx$
$= x e^{x} - e^{x} + C$ $= xe^(x)-e^(x)+C$

$= e^{x} (x - 1) + C$ $=e^(x)(x-1)+C$

In this case, you'd have to perform integration by parts, which is given by the following formula:

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

Note that if you would have chosen $u = e^{x}$ $u=e^(x)$ and $d v = x d x$ $dv = x dx$ , your integral would be even more complicated.

What is the antiderivative of xexxe^x?