int e^x frac{x^2+1}{(x+1)^2} d x is equal to | vedclass

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(C) Let $I = \int e^x \frac{x^2+1}{(x+1)^2} d x$. We can rewrite the numerator as $x^2 - 1 + 2$. $I = \int e^x \left( \frac{x^2-1+2}{(x+1)^2} \right)

Vedclass · Accepted Answer

$e^x\left(\frac{x-1}{x+1}\right)+C$

Vedclass · Answer

(C) Let $I = \int e^x \frac{x^2+1}{(x+1)^2} d x$. We can rewrite the numerator as $x^2 - 1 + 2$. $I = \int e^x \left( \frac{x^2-1+2}{(x+1)^2} \right) d x = \int e^x \left( \frac{(x-1)(x+1)}{(x+1)^2} + \frac{2}{(x+1)^2} \right) d x$. $I = \int e^x \left( \frac{x-1}{x+1} + \frac{2}{(x+1)^2} \right) d x$. Let $f(x) = \frac{x-1}{x+1}$. Then $f'(x) = \frac{(x+1)(1) - (x-1)(1)}{(x+1)^2} = \frac{x+1-x+1}{(x+1)^2} = \frac{2}{(x+1)^2}$. Using the standard integral formula $\int e^x \{f(x) + f'(x)\} d x = e^x f(x) + C$,we get: $I = e^x \left( \frac{x-1}{x+1} \right) + C$.

$\int e^x \frac{x^2+1}{(x+1)^2} d x$ is equal to

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