Find : int frac{(x^{2}+1) e^{x}}{(x+1)^{2}} d | vedclass

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(A) We have $I = \int \frac{(x^{2}+1) e^{x}}{(x+1)^{2}} dx = \int e^{x} \left[ \frac{x^{2}-1+2}{(x+1)^{2}} \right] dx$$= \int e^{x} \left[ \frac{x^{2}

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(A) We have $I = \int \frac{(x^{2}+1) e^{x}}{(x+1)^{2}} dx = \int e^{x} \left[ \frac{x^{2}-1+2}{(x+1)^{2}} \right] dx$
$= \int e^{x} \left[ \frac{x^{2}-1}{(x+1)^{2}} + \frac{2}{(x+1)^{2}} \right] dx = \int e^{x} \left[ \frac{(x-1)(x+1)}{(x+1)^{2}} + \frac{2}{(x+1)^{2}} \right] dx$
$= \int e^{x} \left[ \frac{x-1}{x+1} + \frac{2}{(x+1)^{2}} \right] dx$
Let $f(x) = \frac{x-1}{x+1}$. Then,$f'(x) = \frac{(1)(x+1) - (x-1)(1)}{(x+1)^{2}} = \frac{x+1-x+1}{(x+1)^{2}} = \frac{2}{(x+1)^{2}}$.
Since the integral is of the form $\int e^{x} [f(x) + f'(x)] dx = e^{x} f(x) + C$,
We get $I = e^{x} \left( \frac{x-1}{x+1} \right) + C$.

Find : $\int \frac{(x^{2}+1) e^{x}}{(x+1)^{2}} d x$

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