# Integrate the function $\int\large\frac{e^x}{(1+e^x)(2+e^x)}$

Toolbox:
• $\int \large\frac{1}{(x+a)}$$dx=\log|x+a|+c Step 1: Let I=\int \large\frac{e^x}{(1+e^x)(2+e^x)}$$dx$
$e^x=t$
On differentiating with respect to $x$ we get,
$e^xdx=dt$
$I=\int\large\frac{e^x dt}{(1+t)(2+t)e^x}$
$\;\;\;=\int\large\frac{ dt}{(1+t)(2+t)}$
Step 2:
Let $\large\frac{1}{(1+t)(2+t)}=\frac{A}{(1+t)}+\frac{B}{(2+t)}$-----(1)
$\Rightarrow 1=A(2+t)+B(1+t)$
$\Rightarrow (2A+B)+t(A+B)$
On equating the coefficients of $t$ and constant term on both sides we get
$(A+B)=0$ and $2A+B=1$
Step 3:
On solving both equations we get,
$A+B=0$
$2A+B=1$
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$-A=-1$
$A=1$
$A+B=0$
$1+B=0$
$B=-1$
Step 4:
Substitute the value of A and B in equ(1)
$I=\int \large\frac{1}{1+t}$$dt-\int \large\frac{1}{2+t}$$dt$
Differentiating the above equation with respect to $t$ we get,
$I=\log \mid 1+t\mid -\log \mid 2+t\mid +c$
$\;\;=\log\bigg|\large\frac{1+t}{2+t}\bigg|$$+c t=e^x \;\;=\log\bigg|\large\frac{1+e^x}{2+e^x}\bigg|$$+c$