Question

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

Answer 1

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

let $e^x=t$

then $e^{x}dr=dt$

$I\therefore =\int \frac{1}{(1+t)(2+t)}dt$

now using particle fraction, we can write

$\frac{1}{(1+t)(2+t)}=\frac{A}{1+t}+\frac{B}{2+t}$

$=\frac{A(2+t)+B(1+t)}{(1+t)(2+t)}$

$\frac{(A+B)t+(2A+B)}{(1+t)(2+t)}$

compare numerator at both sides we get

$A+B=0$

and $+2A+B=+1$

$-$ $-$ $-$

$\overline{-A=-1}$

$\therefore A=1$

then $B=-A=-1$

$\therefore B=-1$

So $I\int (\frac{1}{1+t}+\frac{-1}{2+t})dt$

$=\log |1+t|-\log |2+t|+c$

$\log |\frac{1+t}{2+t}|+c$

$\log |\frac{1+e^x}{2+e^x}|+c$