int frac{e^x}{(1 + e^x)(2 + e^x)} dx = | vedclass

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(B) Let $I = \int \frac{e^x}{(1 + e^x)(2 + e^x)} dx$. Using partial fractions,let $\frac{e^x}{(1 + e^x)(2 + e^x)} = \frac{A}{1 + e^x} + \frac{B}{2 + e

Vedclass · Accepted Answer

$\log \left( \frac{1 + e^x}{2 + e^x} \right) + c$

Vedclass · Answer

(B) Let $I = \int \frac{e^x}{(1 + e^x)(2 + e^x)} dx$. Using partial fractions,let $\frac{e^x}{(1 + e^x)(2 + e^x)} = \frac{A}{1 + e^x} + \frac{B}{2 + e^x}$. Then $e^x = A(2 + e^x) + B(1 + e^x)$. Setting $e^x = -1$,we get $A = -1$. Setting $e^x = -2$,we get $B = 1$. So,the integral becomes $\int \left( \frac{1}{2 + e^x} - \frac{1}{1 + e^x} \right) e^x dx$. Let $u = e^x$,then $du = e^x dx$. The integral is $\int \left( \frac{1}{2 + u} - \frac{1}{1 + u} \right) du = \log|2 + u| - \log|1 + u| + c = \log \left| \frac{2 + e^x}{1 + e^x} \right| + c$. Note: The provided option $B$ is $\log \left( \frac{1 + e^x}{2 + e^x} \right) + c$,which is $-\log \left| \frac{2 + e^x}{1 + e^x} \right| + c$. Since the constant $c$ is arbitrary,the result is equivalent to $\log \left| \frac{1 + e^x}{2 + e^x} \right| + c$.

$\int \frac{e^x}{(1 + e^x)(2 + e^x)} dx = $

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