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
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