How do you integrate $\int x e^{x} d x$ $int xe^x dx$ using integration by parts?

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Answer 1 · 2016-01-09T00:41:20

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

$I = \int x e^{x} d x$ $I = intxe^xdx$

Say $u = x$ $u = x$ so $d u = 1$ $du = 1$ , $d v = v = e^{x}$ $dv = v = e^x$

$I = x e^{x} - \int e^{x} d x$ $I = xe^x - inte^xdx$
$I = x e^{x} - e^{x}$ $I = xe^x - e^x$
$I = e^{x} (x - 1) + c$ $I = e^x(x - 1) + c$

How do you integrate ∫xexdxint xe^x dx using integration by parts?