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

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(A) We know that $\int e^x [f(x) + f'(x)] dx = e^x f(x) + c$. Given integral is $I = \int e^x \frac{x^2 + 1}{(x + 1)^2} dx$. Rewrite the numerator: $x

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

Vedclass · Answer

(A) We know that $\int e^x [f(x) + f'(x)] dx = e^x f(x) + c$. Given integral is $I = \int e^x \frac{x^2 + 1}{(x + 1)^2} dx$. Rewrite the numerator: $x^2 + 1 = (x^2 - 1) + 2 = (x - 1)(x + 1) + 2$. So,$I = \int e^x \left[ \frac{(x - 1)(x + 1)}{(x + 1)^2} + \frac{2}{(x + 1)^2} \right] dx$. $I = \int e^x \left[ \frac{x - 1}{x + 1} + \frac{2}{(x + 1)^2} \right] dx$. Let $f(x) = \frac{x - 1}{x + 1}$. Then $f'(x) = \frac{(x + 1)(1) - (x - 1)(1)}{(x + 1)^2} = \frac{x + 1 - x + 1}{(x + 1)^2} = \frac{2}{(x + 1)^2}$. Since the integral is in the form $\int e^x [f(x) + f'(x)] dx$,the result is $e^x f(x) + c$. Thus,$I = e^x \left( \frac{x - 1}{x + 1} \right) + c$.

$\int {e^x \frac{x^2 + 1}{(x + 1)^2} dx} = $

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