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

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(C) We have $I = \int e^x \left(\frac{x+2}{x+4}\right)^2 dx$. Rewrite the numerator as $(x+4-2)$: $I = \int e^x \left(\frac{x+4-2}{x+4}\right)^2 dx =

Vedclass · Accepted Answer

$\frac{x e^x}{(x+4)} + c$

Vedclass · Answer

(C) We have $I = \int e^x \left(\frac{x+2}{x+4}\right)^2 dx$. Rewrite the numerator as $(x+4-2)$: $I = \int e^x \left(\frac{x+4-2}{x+4}\right)^2 dx = \int e^x \left(1 - \frac{2}{x+4}\right)^2 dx$. Expanding the square: $I = \int e^x \left(1 - \frac{4}{x+4} + \frac{4}{(x+4)^2}\right) dx$. This can be written as: $I = \int e^x \left(1 - \frac{4}{x+4}\right) dx + \int \frac{4 e^x}{(x+4)^2} dx$. Let $f(x) = 1 - \frac{4}{x+4}$. Then $f'(x) = -(-4)(x+4)^{-2} = \frac{4}{(x+4)^2}$. Using the standard integral formula $\int e^x (f(x) + f'(x)) dx = e^x f(x) + c$: $I = e^x \left(1 - \frac{4}{x+4}\right) + c = e^x \left(\frac{x+4-4}{x+4}\right) + c = \frac{x e^x}{x+4} + c$.

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

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