Question:

Step 1: Rewrite the integrand by dividing the numerator and denominator by $e^{2x}$. To make a suitable substitution, we transform the integrand by multiplying the numerator and denominator by $e^{-2x}$. $$I = \int \frac{x^2 + x}{(e^x + x + 1)^2} dx = \int \frac{(x^2 + x)e^{-2x}}{(e^x(1 + xe^{-x} + e^{-x}))^2} dx$$ $$I = \int \frac{(x^2 + x)e^{-2x}}{e^{2x}(1 + xe^{-x} + e^{-x})^2} dx = \int \frac{(x^2 + x)e^{-2x}}{(1 + (x+1)e^{-x})^2} dx$$ Step 2: Apply a suitable substitution. Let $u = 1 + (x+1)e^{-x}$. We need to find $du$ in terms of $dx$. Differentiating $u$ with respect to $x$: $$du = \frac{d}{dx} (1 + (x+1)e^{-x}) dx$$ Applying the product rule for $(x+1)e^{-x}$: $$du = \left( (1 \cdot e^{-x}) + ((x+1) \cdot (-e^{-x})) \right) dx$$ $$du = (e^{-x} - (x+1)e^{-x}) dx$$ $$du = (e^{-x} - xe^{-x} - e^{-x}) dx$$ $$du = -xe^{-x} dx$$ Step 3: Express the numerator in terms of $u$ and $du$. The numerator of the rewritten integral is $(x^2 + x)e^{-2x} dx$. We can factor out $x$ from $x^2+x$ and group terms: $$(x^2 + x)e^{-2x} dx = x(x+1)e^{-2x} dx$$ This can be written as $(x+1)e^{-x} \cdot (xe^{-x}) dx$. From our substitution: $$(x+1)e^{-x} = u-1$$ And from the differential $du$: $$xe^{-x} dx = -du$$ Substituting these into the numerator expression: $$(x+1)e^{-x} \cdot (xe^{-x}) dx = (u-1)(-du) = (1-u)du$$ Step 4: Substitute into the integral and evaluate. Now, substitute $u$ and $du$ into the integral: $$I = \int \frac{(1-u)du}{u^2}$$ Split the integrand into simpler terms: $$I = \int \left(\frac{1}{u^2} - \frac{u}{u^2}\right) du = \int \left(\frac{1}{u^2} - \frac{1}{u}\right) du$$ Integrate term by term: $$I = -\frac{1}{u} - \ln|u| + C$$ Step 5: Substitute back to express the result in terms of $x$. Substitute $u = 1 + (x+1)e^{-x}$ back into the result: $$I = -\frac{1}{1 + (x+1)e^{-x}} - \ln|1 + (x+1)e^{-x}| + C$$ Step 6: State the final answer. The final integral value is $C - \ln|1 + (x+1)e^{-x}| - \frac{1}{1 + (x+1)e^{-x}}$. This matches Option 4. The final answer is $\boxed{\text{Option 4}}$.