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

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Question

(C) Let $I = \int \frac{x+1}{x(1+x e^x)^2} \,d x$.
Multiply the numerator and denominator by $e^x$:
$I = \int \frac{(x+1)e^x}{x e^x(1+x e^x)^2} \,d

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Let $I = \int \frac{x+1}{x(1+x e^x)^2} \,d x$.
Multiply the numerator and denominator by $e^x$:
$I = \int \frac{(x+1)e^x}{x e^x(1+x e^x)^2} \,d x$.
Let $t = x e^x$. Then $dt = (e^x + x e^x) dx = e^x(1+x) dx$.
Substituting these into the integral:
$I = \int \frac{dt}{t(1+t)^2}$.
Using partial fractions: $\frac{1}{t(1+t)^2} = \frac{A}{t} + \frac{B}{1+t} + \frac{C}{(1+t)^2}$.
Solving for constants,we get $A=1, B=-1, C=-1$.
$I = \int \left( \frac{1}{t} - \frac{1}{1+t} - \frac{1}{(1+t)^2} \right) dt$.
$I = \log|t| - \log|1+t| + \frac{1}{1+t} + c$.
Substituting $t = x e^x$ back:
$I = \log \left| \frac{x e^x}{1+x e^x} \right| + \frac{1}{1+x e^x} + c$.

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

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