1. **Problem:** Calculate the integral $$\int \frac{x^2}{(1-x)^4} \, dx$$. 2. **Formula and approach:** We will use substitution and integration by parts if needed. Important rule: when the denominator is a power of a binomial, substitution $u = 1 - x$ often simplifies the integral. 3. **Step 1: Substitution** Let $$u = 1 - x \implies x = 1 - u$$ and $$dx = -du$$. Rewrite the integral: $$\int \frac{x^2}{(1-x)^4} dx = \int \frac{(1-u)^2}{u^4} (-du) = -\int \frac{(1-u)^2}{u^4} du$$ 4. **Step 2: Expand numerator** $$(1-u)^2 = 1 - 2u + u^2$$ So the integral becomes: $$-\int \frac{1 - 2u + u^2}{u^4} du = -\int \left(u^{-4} - 2u^{-3} + u^{-2}\right) du$$ 5. **Step 3: Integrate term by term** $$-\left( \int u^{-4} du - 2 \int u^{-3} du + \int u^{-2} du \right)$$ Calculate each: - $$\int u^{-4} du = \frac{u^{-3}}{-3} = -\frac{1}{3u^3}$$ - $$\int u^{-3} du = \frac{u^{-2}}{-2} = -\frac{1}{2u^2}$$ - $$\int u^{-2} du = \frac{u^{-1}}{-1} = -\frac{1}{u}$$ 6. **Step 4: Substitute back and simplify** $$-\left(-\frac{1}{3u^3} - 2 \left(-\frac{1}{2u^2}\right) - \frac{1}{u}\right) + C = -\left(-\frac{1}{3u^3} + \frac{1}{u^2} - \frac{1}{u}\right) + C$$ $$= \frac{1}{3u^3} - \frac{1}{u^2} + \frac{1}{u} + C$$ Recall $$u = 1 - x$$, so: $$\boxed{\int \frac{x^2}{(1-x)^4} dx = \frac{1}{3(1-x)^3} - \frac{1}{(1-x)^2} + \frac{1}{1-x} + C}$$