int frac{(x + 1)}{x(1 + xe^x)^2} dx = | vedclass

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(B) Multiply the numerator and denominator by $e^x$: $I = \int \frac{(x + 1)e^x}{xe^x(1 + xe^x)^2} dx$ Let $u = xe^x$, then $du = (e^x + xe^x) dx = e^

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$\log \left( \frac{xe^x}{1 + xe^x} \right) + \frac{1}{(1 + xe^x)} + c$

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(B) Multiply the numerator and denominator by $e^x$: $I = \int \frac{(x + 1)e^x}{xe^x(1 + xe^x)^2} dx$ Let $u = xe^x$, then $du = (e^x + xe^x) dx = e^x(1 + x) dx$. Substituting these into the integral: $I = \int \frac{du}{u(1 + u)^2}$ Using partial fractions: $\frac{1}{u(1 + u)^2} = \frac{A}{u} + \frac{B}{1 + u} + \frac{C}{(1 + u)^2}$ $1 = A(1 + u)^2 + Bu(1 + u) + Cu$ For $u = 0$, $A = 1$. For $u = -1$, $C = -1$. Comparing coefficients of $u^2$: $A + B = 0 \implies B = -1$. $I = \int \left( \frac{1}{u} - \frac{1}{1 + u} - \frac{1}{(1 + u)^2} \right) du$ $I = \log|u| - \log|1 + u| + \frac{1}{1 + u} + c$ $I = \log \left| \frac{u}{1 + u} \right| + \frac{1}{1 + u} + c$ Substituting $u = xe^x$: $I = \log \left( \frac{xe^x}{1 + xe^x} \right) + \frac{1}{1 + xe^x} + c$

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

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