Mathematics

$\displaystyle\int \:e^x\left(x^2-x+1\right)dx$

SOLUTION
$\displaystyle\int{{e}^{x}\left({x}^{2}-x+1\right)dx}$
$=\displaystyle\int{{x}^{2}{e}^{x}dx}-\int{x{e}^{x}dx}+\int{{e}^{x}dx}$
Consider $\displaystyle\int{{x}^{2}{e}^{x}dx}$ is of the form $\displaystyle\int{u.dv}=uv-\int{v.du}$
Take $u={x}^{2}\Rightarrow du=2x dx$ and $dv={e}^{x} dx\Rightarrow v={e}^{x}$
$\displaystyle\int{{x}^{2}{e}^{x}dx}-\int{x{e}^{x}dx}+\int{{e}^{x}dx}={x}^{2}{e}^{x}-2\int{x{e}^{x}dx}-\int{x{e}^{x}dx}+\int{{e}^{x}dx}$
$={x}^{2}{e}^{x}-3\int{x{e}^{x}dx}+\int{{e}^{x}dx}$
Consider $\displaystyle\int{x{e}^{x}dx}$
Take $u=x\Rightarrow du=dx$ and $dv={e}^{x}dx\Rightarrow v={e}^{x}$
$\Rightarrow {x}^{2}{e}^{x}-3\displaystyle\int{x{e}^{x}dx}+\int{{e}^{x}dx}={x}^{2}{e}^{x}-3\left(x{e}^{x}-\int{{e}^{x}dx}\right)+\int{{e}^{x}dx}$
$={x}^{2}{e}^{x}-3x{e}^{x}+4\displaystyle\int{{e}^{x}dx}$
$={x}^{2}{e}^{x}-3x{e}^{x}+4{e}^{x}$
$=\left({x}^{2}-3x+4\right){e}^{x}+c$ where $c$ is the constant of integration.

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
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