1. Substitution: Let \( u = xe^x \). Then, the derivative of \( u \) with respect to \( x \) is: \[ \frac{du}{dx} = e^x + xe^x = e^x (1 + x) \] Therefore: \[ du = e^x (1 + x) \, dx \] Notice that \( x + 1 = \frac{du}{e^x dx} \). 2. Rewrite the Integral: Substitute \( u = xe^x \) and \( du = e^x (1 + x) \, dx \) into the integral: \[ \int \frac{x+1}{x(xe^x + 1)} \, dx = \int \frac{1}{x} \cdot \frac{x+1}{xe^x + 1} \, dx \] Using \( u = xe^x \), we have: \[ \int \frac{1}{x} \cdot \frac{du}{e^x dx} \cdot \frac{1}{u + 1} \, dx = \int \frac{1}{u(u + 1)} \, du \] 3. Partial Fraction Decomposition: Decompose \( \frac{1}{u(u + 1)} \) into partial fractions: \[ \frac{1}{u(u + 1)} = \frac{1}{u} - \frac{1}{u + 1} \] Therefore: \[ \int \frac{1}{u(u + 1)} \, du = \int \left( \frac{1}{u} - \frac{1}{u + 1} \right) \, du \] 4. Integrate: Integrate each term separately: \[ \int \frac{1}{u} \, du - \int \frac{1}{u + 1} \, du = \ln |u| - \ln |u + 1| + C \] Combine the logarithms: \[ \ln \left( \frac{u}{u + 1} \right) + C \] 5. Substitute Back \( u = xe^x \): Replace \( u \) with \( xe^x \): \[ \ln \left( \frac{xe^x}{xe^x + 1} \right) + C \] Therefore, the value of the integral is: \[ \boxed{\ln \left( \frac{xe^x}{1 + xe^x} \right) + C} \]