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

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(D) Let $I = \int \frac{x+1}{x(1+x e^x)^2} dx$. Multiply numerator and denominator by $e^x$: $I = \int \frac{e^x(x+1)}{x e^x(1+x e^x)^2} dx$. Let $t =

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$\log \left|\frac{x e^x}{1+x e^x}\right|+\frac{1}{1+x e^x}+c$,where $c$ is a constant of integration.

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(D) Let $I = \int \frac{x+1}{x(1+x e^x)^2} dx$. Multiply numerator and denominator by $e^x$: $I = \int \frac{e^x(x+1)}{x e^x(1+x e^x)^2} dx$. Let $t = x e^x$. Then $dt = (e^x + x e^x) dx = e^x(1+x) dx$. Substituting this 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: $1 = A(1+t)^2 + Bt(1+t) + Ct$. For $t=0$,$A=1$. For $t=-1$,$C=-1$. Comparing coefficients of $t^2$,$A+B=0 \Rightarrow B=-1$. So,$I = \int \left( \frac{1}{t} - \frac{1}{1+t} - \frac{1}{(1+t)^2} \right) dt$. $I = \log |t| - \log |1+t| + \frac{1}{1+t} + c$. Substituting $t = x e^x$: $I = \log |x e^x| - \log |1+x e^x| + \frac{1}{1+x e^x} + c$. $I = \log \left| \frac{x e^x}{1+x e^x} \right| + \frac{1}{1+x e^x} + c$.

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

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