int left(frac{x+2}{x+4}right)^2 e^x , dx = | vedclass

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Question

(A) We have $I = \int \left(\frac{x+2}{x+4}\right)^2 e^x \, dx$. First,rewrite the integrand: $I = \int \left(\frac{(x+4)-2}{x+4}\right)^2 e^x \, dx =

Vedclass · Accepted Answer

$e^x \left(\frac{x}{x+4}\right) + c$,where $c$ is a constant of integration.

Vedclass · Answer

(A) We have $I = \int \left(\frac{x+2}{x+4}\right)^2 e^x \, dx$. First,rewrite the integrand: $I = \int \left(\frac{(x+4)-2}{x+4}\right)^2 e^x \, dx = \int \left(1 - \frac{2}{x+4}\right)^2 e^x \, dx$. Expanding the square: $I = \int \left(1 - \frac{4}{x+4} + \frac{4}{(x+4)^2}\right) e^x \, dx$. Let $f(x) = 1 - \frac{4}{x+4} = \frac{x+4-4}{x+4} = \frac{x}{x+4}$. Then $f'(x) = \frac{d}{dx} \left(1 - 4(x+4)^{-1}\right) = 0 - 4(-1)(x+4)^{-2} = \frac{4}{(x+4)^2}$. Since the integral is of the form $\int e^x (f(x) + f'(x)) \, dx = e^x f(x) + c$,$I = e^x \left(\frac{x}{x+4}\right) + c$.

$\int \left(\frac{x+2}{x+4}\right)^2 e^x \, dx =$

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