The value of int frac{x+1}{x(1+x e^x)^2} ,dx | vedclass

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(C) 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

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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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(C) 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 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 $A=1, B=-1, C=-1$. 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 $t = x e^x$ back: $I = \log \left| \frac{x e^x}{1+x e^x} \right| + \frac{1}{1+x e^x} + c$.

The value of $\int \frac{x+1}{x(1+x e^x)^2} \,dx$ is equal to

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