∫((x^2 +1)/((x+1)^2 ))e^x dx


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Question Number 69715 by Mikaell last updated on 26/Sep/19

$\int \frac{x^{2} + 1}{{(x + 1)}^{2}} e^{x} d x$
	Answered by mind is power last updated on 27/Sep/19

	$\int \frac{x^{2} + 1}{x^{2} + 2 x + 1} e^{x} d x = \int \frac{x^{2} + 2 x + 1 - 2 x}{(x^{2} + 2 x + 1)} = \int e^{x} d x - \int \frac{2 x}{x^{2} + 2 x + 1} e^{x} d x$ $\int e^{x} d x - 2 \int \frac{(x + 1) e^{x} - e^{x}}{{(x + 1)}^{2}} d x = \int e^{x} d x - 2 \int \frac{(x + 1) {(e^{x})}^{'} - {(x + 1)}^{'} e^{x}}{{(x + 1)}^{2}} \int e^{x} d x - 2 \int d (\frac{e^{x}}{x + 1})$ $= e^{x} - 2 \frac{e^{x}}{x + 1} + c = \frac{(x - 1) e^{x}}{x + 1} + c$
	Commented by Mikaell last updated on 27/Sep/19

	$t h a n k y o u S i r$
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