int frac{(x^{2}+1) e^{x}}{(x+1)^{2}} d x=f(x) | vedclass

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(A) We have $\int \frac{(x^{2}+1) e^{x}}{(x+1)^{2}} d x = \int \frac{(x^{2}-1+2) e^{x}}{(x+1)^{2}} d x$.$= \int \left( \frac{(x-1)(x+1)}{(x+1)^{2}} +

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$\frac{3}{4}$

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(A) We have $\int \frac{(x^{2}+1) e^{x}}{(x+1)^{2}} d x = \int \frac{(x^{2}-1+2) e^{x}}{(x+1)^{2}} d x$.$= \int \left( \frac{(x-1)(x+1)}{(x+1)^{2}} + \frac{2}{(x+1)^{2}} \right) e^{x} d x = \int \left( \frac{x-1}{x+1} + \frac{2}{(x+1)^{2}} \right) e^{x} d x$.This is in the form $\int (f(x) + f'(x)) e^{x} d x = f(x) e^{x} + C$,where $f(x) = \frac{x-1}{x+1}$.We can write $f(x) = \frac{x+1-2}{x+1} = 1 - 2(x+1)^{-1}$.Then $f'(x) = 2(x+1)^{-2}$.$f''(x) = -4(x+1)^{-3}$.$f'''(x) = 12(x+1)^{-4} = \frac{12}{(x+1)^{4}}$.At $x = 1$,$f'''(1) = \frac{12}{(1+1)^{4}} = \frac{12}{16} = \frac{3}{4}$.

$\int \frac{(x^{2}+1) e^{x}}{(x+1)^{2}} d x=f(x) e^{x}+C$,where $C$ is a constant,then $\frac{d^{3} f}{d x^{3}}$ at $x = 1$ is equal to

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