1. **Problem statement:** Evaluate the integral $$\int \frac{2 + x}{(1 + x)^2} \, dx$$. 2. **Formula and approach:** We can split the integral into simpler parts or use substitution. Here, let’s split the numerator: $$\int \frac{2 + x}{(1 + x)^2} \, dx = \int \frac{2}{(1 + x)^2} \, dx + \int \frac{x}{(1 + x)^2} \, dx$$. 3. **Evaluate the first integral:** Use substitution $u = 1 + x$, so $du = dx$. $$\int \frac{2}{u^2} \, du = 2 \int u^{-2} \, du = 2 \left(-u^{-1}\right) + C = -\frac{2}{1 + x} + C$$. 4. **Evaluate the second integral:** Rewrite $x = u - 1$, so $$\int \frac{x}{(1 + x)^2} \, dx = \int \frac{u - 1}{u^2} \, du = \int \left(\frac{u}{u^2} - \frac{1}{u^2}\right) du = \int \left(u^{-1} - u^{-2}\right) du$$. 5. **Integrate term by term:** $$\int u^{-1} \, du = \ln|u| + C$$ $$\int u^{-2} \, du = -u^{-1} + C$$ So, $$\int \left(u^{-1} - u^{-2}\right) du = \ln|u| + \frac{1}{u} + C = \ln|1 + x| + \frac{1}{1 + x} + C$$. 6. **Combine results:** $$\int \frac{2 + x}{(1 + x)^2} \, dx = -\frac{2}{1 + x} + \ln|1 + x| + \frac{1}{1 + x} + C = \ln|1 + x| - \frac{1}{1 + x} + C$$. --- 1. **Problem statement:** Evaluate the integral $$\int \frac{2x^2}{1 - x^2} \, dx$$. 2. **Rewrite the integrand:** Note that $1 - x^2 = (1 - x)(1 + x)$, but better to use polynomial division since degree numerator $=2$ and denominator $=2$. 3. **Divide numerator by denominator:** $$\frac{2x^2}{1 - x^2} = \frac{2x^2}{1 - x^2} = \frac{2x^2}{-(x^2 - 1)} = -\frac{2x^2}{x^2 - 1}$$. Divide $2x^2$ by $x^2 - 1$: $$2x^2 \div (x^2 - 1) = 2 + \frac{2}{x^2 - 1}$$. So, $$-\frac{2x^2}{x^2 - 1} = -(2 + \frac{2}{x^2 - 1}) = -2 - \frac{2}{x^2 - 1}$$. 4. **Rewrite integral:** $$\int \frac{2x^2}{1 - x^2} \, dx = \int \left(-2 - \frac{2}{x^2 - 1}\right) dx = -2 \int dx - 2 \int \frac{1}{x^2 - 1} dx$$. 5. **Integrate first term:** $$-2 \int dx = -2x + C$$. 6. **Integrate second term:** Use partial fractions: $$\frac{1}{x^2 - 1} = \frac{1}{(x - 1)(x + 1)} = \frac{A}{x - 1} + \frac{B}{x + 1}$$. Solve for $A$ and $B$: $$1 = A(x + 1) + B(x - 1)$$. Set $x=1$: $$1 = A(2) + B(0) \Rightarrow A = \frac{1}{2}$$. Set $x=-1$: $$1 = A(0) + B(-2) \Rightarrow B = -\frac{1}{2}$$. So, $$\int \frac{1}{x^2 - 1} dx = \int \left(\frac{1/2}{x - 1} - \frac{1/2}{x + 1}\right) dx = \frac{1}{2} \ln|x - 1| - \frac{1}{2} \ln|x + 1| + C$$. 7. **Combine logarithms:** $$\frac{1}{2} \ln \left|\frac{x - 1}{x + 1}\right| + C$$. 8. **Final answer:** $$\int \frac{2x^2}{1 - x^2} dx = -2x - 2 \cdot \frac{1}{2} \ln \left|\frac{x - 1}{x + 1}\right| + C = -2x - \ln \left|\frac{x - 1}{x + 1}\right| + C$$.