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

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(C) Let $t = e^x$,then $dt = e^x dx$. Substituting this into the integral,we get: $\int \frac{dt}{(t+2)(t+1)}$. Using partial fractions: $\frac{1}{(t+

Vedclass · Accepted Answer

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

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(C) Let $t = e^x$,then $dt = e^x dx$. Substituting this into the integral,we get: $\int \frac{dt}{(t+2)(t+1)}$. Using partial fractions: $\frac{1}{(t+2)(t+1)} = \frac{A}{t+2} + \frac{B}{t+1}$. $1 = A(t+1) + B(t+2)$. For $t = -1$,$B = 1$. For $t = -2$,$A = -1$. So,$\int (\frac{-1}{t+2} + \frac{1}{t+1}) dt = -\log|t+2| + \log|t+1| + C$. $= \log|\frac{t+1}{t+2}| + C$. Substituting $t = e^x$ back,we get: $\log \left(\frac{e^x+1}{e^x+2}\right) + C$.

$\int \frac{e^x}{(2+e^x)(e^x+1)} dx =$ (where $C$ is a constant of integration.)

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