Mathematics

Evaluate : $$\displaystyle \int \dfrac{(x^{2}+1)e^{x}}{(x+1)^{2}} dx$$


SOLUTION
Let $$I=\displaystyle\int{{e}^{x}\left(\dfrac{{x}^{2}+1}{{\left(x+1\right)}^{2}}\right)dx}$$
$$=\displaystyle\int{{e}^{x}\left(\dfrac{{x}^{2}-1+2}{{\left(x+1\right)}^{2}}\right)dx}$$
$$=\displaystyle\int{{e}^{x}\left(\dfrac{\left(x-1\right)\left(x+1\right)+2}{{\left(x+1\right)}^{2}}\right)dx}$$
$$=\displaystyle\int{{e}^{x}\dfrac{x-1}{x+1}}+\displaystyle\int{\dfrac{2{e}^{x}}{{\left(x+1\right)}^{2}}}$$
$$=\left(\dfrac{x-1}{x+1}\right)\times \displaystyle\int{{e}^{x}dx}-\displaystyle\int{\left[\dfrac{d}{dx}\left(\dfrac{x-1}{x+1}\right)\times \int{{e}^{x}dx}\right]dx}+2\displaystyle\int{\dfrac{{e}^{x}dx}{{\left(x+1\right)}^{2}}}$$
$$=\left(\dfrac{x-1}{x+1}\right){e}^{x}- \displaystyle\int{\left[\dfrac{\left(x+1\right)-\left(x-1\right)\times 1}{{\left(x+1\right)}^{2}}{e}^{x}\right]dx}+2\displaystyle\int{\dfrac{{e}^{x}dx}{{\left(x+1\right)}^{2}}}$$
$$=\left(\dfrac{x-1}{x+1}\right){e}^{x}-2\displaystyle\int{\dfrac{{e}^{x}dx}{{\left(x+1\right)}^{2}}}+2\displaystyle\int{\dfrac{{e}^{x}dx}{{\left(x+1\right)}^{2}}}$$
$$={e}^{x}\left(\dfrac{x-1}{x+1}\right)+c$$

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Subjective Medium Published on 17th 09, 2020
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