1. **Problem Statement:** Evaluate the integral $$\int \frac{x^3 + x - 1}{(x^2 + 1)^2} \, dx.$$ 2. **Formula and Rules:** For integrals involving repeated quadratic factors like $$(x^2 + 1)^2,$$ we use partial fraction decomposition or substitution methods. Important to note: - The denominator is a repeated quadratic factor. - Numerator degree is less than denominator degree after expansion, so decomposition is possible. 3. **Step 1: Rewrite the integral** $$\int \frac{x^3 + x - 1}{(x^2 + 1)^2} \, dx = \int \frac{x^3 + x - 1}{(x^2 + 1)^2} \, dx.$$ 4. **Step 2: Express numerator in terms of $x^2 + 1$ and $x$:** Try to write numerator as $$A x + B + C (x^2 + 1)$$ to simplify. But since numerator is cubic, try polynomial division first. 5. **Step 3: Polynomial division:** Divide numerator $x^3 + x - 1$ by denominator base $x^2 + 1$: - Quotient: $x$ - Remainder: $x - 1 - x = -1$ So, $$\frac{x^3 + x - 1}{(x^2 + 1)^2} = \frac{x(x^2 + 1) - 1}{(x^2 + 1)^2} = \frac{x}{x^2 + 1} - \frac{1}{(x^2 + 1)^2}.$$ 6. **Step 4: Split the integral:** $$\int \frac{x^3 + x - 1}{(x^2 + 1)^2} \, dx = \int \frac{x}{x^2 + 1} \, dx - \int \frac{1}{(x^2 + 1)^2} \, dx.$$ 7. **Step 5: Evaluate each integral separately:** - First integral: $$\int \frac{x}{x^2 + 1} \, dx.$$ Use substitution: let $$u = x^2 + 1,$$ then $$du = 2x \, dx,$$ so $$x \, dx = \frac{du}{2}.$$ Therefore, $$\int \frac{x}{x^2 + 1} \, dx = \int \frac{1}{u} \cdot \frac{du}{2} = \frac{1}{2} \int \frac{1}{u} \, du = \frac{1}{2} \ln|u| + C = \frac{1}{2} \ln(x^2 + 1) + C.$$ - Second integral: $$\int \frac{1}{(x^2 + 1)^2} \, dx.$$ This is a standard integral with formula: $$\int \frac{dx}{(x^2 + a^2)^2} = \frac{x}{2a^2(x^2 + a^2)} + \frac{1}{2a^3} \arctan\left(\frac{x}{a}\right) + C.$$ For $$a = 1,$$ $$\int \frac{dx}{(x^2 + 1)^2} = \frac{x}{2(x^2 + 1)} + \frac{1}{2} \arctan(x) + C.$$ 8. **Step 6: Combine results:** $$\int \frac{x^3 + x - 1}{(x^2 + 1)^2} \, dx = \frac{1}{2} \ln(x^2 + 1) - \left( \frac{x}{2(x^2 + 1)} + \frac{1}{2} \arctan(x) \right) + C.$$ Simplify: $$= \frac{1}{2} \ln(x^2 + 1) - \frac{x}{2(x^2 + 1)} - \frac{1}{2} \arctan(x) + C.$$