int left( frac{x^2+1}{(x+1)^2} right) e^x , d | vedclass

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(A) We know that $\int e^x [f(x) + f'(x)] \, dx = e^x f(x) + c$. First,rewrite the integrand: $\frac{x^2+1}{(x+1)^2} = \frac{x^2-1+2}{(x+1)^2} = \frac

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) We know that $\int e^x [f(x) + f'(x)] \, dx = e^x f(x) + c$. First,rewrite the integrand: $\frac{x^2+1}{(x+1)^2} = \frac{x^2-1+2}{(x+1)^2} = \frac{(x-1)(x+1)+2}{(x+1)^2} = \frac{x-1}{x+1} + \frac{2}{(x+1)^2}$. Let $f(x) = \frac{x-1}{x+1}$. Then $f'(x) = \frac{(x+1)(1) - (x-1)(1)}{(x+1)^2} = \frac{x+1-x+1}{(x+1)^2} = \frac{2}{(x+1)^2}$. Thus,the integral becomes $\int e^x [f(x) + f'(x)] \, dx = e^x f(x) + c = \left( \frac{x-1}{x+1} \right) e^x + c$.

$\int \left( \frac{x^2+1}{(x+1)^2} \right) e^x \, dx = \text{ . . . . . . }$.

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