∫e^x(x^2 x+1)dx =∫x^2e^xdx ∫xe^xdx+∫e^xdx Consider ∫x^2e^xdx is of the form ∫u.dv=uv ∫v.du Take u=x^2⇒ du=2x dx and dv=e^x dx⇒ v ...

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Integration by Parts

∫ exx2-x+1dx

Question

$\int e^{x} (x^{2} - x + 1) d x$

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Solution

$\int e^{x} (x^{2} - x + 1) d x$
$= \int x^{2} e^{x} d x - \int x e^{x} d x + \int e^{x} d x$
Consider $\int x^{2} e^{x} d x$ is of the form $\int u . d v = u v - \int v . d u$
Take $u = x^{2} \Rightarrow d u = 2 x d x$ and $d v = e^{x} d x \Rightarrow v = e^{x}$
$\int x^{2} e^{x} d x - \int x e^{x} d x + \int e^{x} d x = x^{2} e^{x} - 2 \int x e^{x} d x - \int x e^{x} d x + \int e^{x} d x$
$= x^{2} e^{x} - 3 \int x e^{x} d x + \int e^{x} d x$
Consider $\int x e^{x} d x$
Take $u = x \Rightarrow d u = d x$ and $d v = e^{x} d x \Rightarrow v = e^{x}$
$\Rightarrow x^{2} e^{x} - 3 \int x e^{x} d x + \int e^{x} d x = x^{2} e^{x} - 3 (x e^{x} - \int e^{x} d x) + \int e^{x} d x$
$= x^{2} e^{x} - 3 x e^{x} + 4 \int e^{x} d x$
$= x^{2} e^{x} - 3 x e^{x} + 4 e^{x}$
$= (x^{2} - 3 x + 4) e^{x} + c$ where $c$ is the constant of integration.

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