int frac{1+2 e^{-x}}{1-2 e^{-x}} d x= | vedclass

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(C) Let $I = \int \frac{1+2 e^{-x}}{1-2 e^{-x}} d x$. We can rewrite the numerator as $(1-2e^{-x}) + 4e^{-x}$. So,$I = \int \frac{(1-2e^{-x}) + 4e^{-x

Vedclass · Accepted Answer

$x+2\log \left|1-2 e^{-x}\right|+c$

Vedclass · Answer

(C) Let $I = \int \frac{1+2 e^{-x}}{1-2 e^{-x}} d x$. We can rewrite the numerator as $(1-2e^{-x}) + 4e^{-x}$. So,$I = \int \frac{(1-2e^{-x}) + 4e^{-x}}{1-2e^{-x}} d x$. $I = \int \left( 1 + \frac{4e^{-x}}{1-2e^{-x}} \right) d x$. $I = \int 1 d x + 2 \int \frac{2e^{-x}}{1-2e^{-x}} d x$. Let $u = 1-2e^{-x}$,then $du = 2e^{-x} d x$. $I = x + 2 \int \frac{1}{u} du = x + 2 \ln|u| + c$. Substituting back $u$,we get $I = x + 2 \ln|1-2e^{-x}| + c$.

$\int \frac{1+2 e^{-x}}{1-2 e^{-x}} d x=$

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