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$\int \frac{e^x}{\left(1+e^x\right)\left(2+e^x\right)} d x=$
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Verified Answer
The correct answer is:
$\log \left[\frac{1+e^x}{2+e^x}\right]+c$
$\int \frac{e^x}{\left(1+e^x\right)\left(2+e^x\right)} d x=\int\left\{\frac{e^x}{1+e^x}-\frac{e^x}{2+e^x}\right\} d x$
Now put $1+e^x=t$ and $2+e^x=t$, then the required integral
$=\log \left(1+e^x\right)-\log \left(2+e^x\right)=\log \left(\frac{1+e^x}{2+e^x}\right)+c .$
Now put $1+e^x=t$ and $2+e^x=t$, then the required integral
$=\log \left(1+e^x\right)-\log \left(2+e^x\right)=\log \left(\frac{1+e^x}{2+e^x}\right)+c .$
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