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

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Question

(A) We are given the integral $I = \int e^x \left(\frac{x+2}{x+4}\right)^2 dx$. First,rewrite the expression inside the square: $\frac{x+2}{x+4} = \fr

Vedclass · Accepted Answer

$\frac{x e^x}{x+4}$

Vedclass · Answer

(A) We are given the integral $I = \int e^x \left(\frac{x+2}{x+4}\right)^2 dx$. First,rewrite the expression inside the square: $\frac{x+2}{x+4} = \frac{x+4-2}{x+4} = 1 - \frac{2}{x+4}$. Thus,$\left(\frac{x+2}{x+4}\right)^2 = \left(1 - \frac{2}{x+4}\right)^2 = 1 - \frac{4}{x+4} + \frac{4}{(x+4)^2}$. Now,the integral becomes: $I = \int e^x \left(1 - \frac{4}{x+4} + \frac{4}{(x+4)^2}\right) dx$. We know the standard form $\int e^x [g(x) + g'(x)] dx = e^x g(x) + C$. Let $g(x) = 1 - \frac{4}{x+4}$. Then $g'(x) = -\left(-\frac{4}{(x+4)^2}\right) = \frac{4}{(x+4)^2}$. Therefore,$I = e^x \left(1 - \frac{4}{x+4}\right) + C = e^x \left(\frac{x+4-4}{x+4}\right) + C = \frac{x e^x}{x+4} + C$. Hence,$f(x) = \frac{x e^x}{x+4}$.

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

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