1. **Problem Statement:** Evaluate the definite integral $$\int_0^1 \frac{x e^x}{(x+1)^2} \, dx$$. 2. **Method:** Use substitution to simplify the integral. Let $$u = x + 1$$, so $$x = u - 1$$ and $$dx = du$$. 3. **Change limits:** When $$x=0$$, $$u=1$$; when $$x=1$$, $$u=2$$. 4. **Rewrite the integral:** $$\int_0^1 \frac{x e^x}{(x+1)^2} \, dx = \int_1^2 \frac{(u-1) e^{u-1}}{u^2} \, du$$. 5. **Simplify the integrand:** $$= \int_1^2 \frac{(u-1)}{u^2} e^{u-1} \, du = \int_1^2 \left(\frac{u}{u^2} - \frac{1}{u^2}\right) e^{u-1} \, du = \int_1^2 \left(\frac{1}{u} - \frac{1}{u^2}\right) e^{u-1} \, du$$. 6. **Rewrite as:** $$\int_1^2 \frac{e^{u-1}}{u} \, du - \int_1^2 \frac{e^{u-1}}{u^2} \, du$$. 7. **Use integration by parts on the second integral:** Let $$I = \int_1^2 \frac{e^{u-1}}{u} \, du$$ and $$J = \int_1^2 \frac{e^{u-1}}{u^2} \, du$$. 8. For $$J$$, set $$v = e^{u-1}$$ and $$w = -\frac{1}{u}$$, then $$J = \int e^{u-1} u^{-2} du = -\frac{e^{u-1}}{u} \Big|_1^2 + \int_1^2 \frac{e^{u-1}}{u} du = -\frac{e^{1}}{2} + \frac{e^{0}}{1} + I = -\frac{e}{2} + 1 + I$$. 9. Substitute back into the original integral: $$I - J = I - \left(-\frac{e}{2} + 1 + I\right) = I - (-\frac{e}{2} + 1 + I) = \frac{e}{2} - 1$$. 10. **Final answer:** $$\int_0^1 \frac{x e^x}{(x+1)^2} \, dx = \frac{e}{2} - 1$$.