1. **State the problem:** We want to find the indefinite integral $$\int \frac{x^4 - x^3 + 2x^2 - x + 2}{(x - 1)(x^2 + 2)^2} \, dx$$ 2. **Formula and approach:** For rational functions, we use **partial fraction decomposition** to express the integrand as a sum of simpler fractions that are easier to integrate. 3. **Set up partial fractions:** Since the denominator is $(x - 1)(x^2 + 2)^2$, the decomposition form is: $$\frac{x^4 - x^3 + 2x^2 - x + 2}{(x - 1)(x^2 + 2)^2} = \frac{A}{x - 1} + \frac{Bx + C}{x^2 + 2} + \frac{Dx + E}{(x^2 + 2)^2}$$ 4. **Multiply both sides by the denominator:** $$x^4 - x^3 + 2x^2 - x + 2 = A(x^2 + 2)^2 + (Bx + C)(x - 1)(x^2 + 2) + (Dx + E)(x - 1)$$ 5. **Expand terms:** - Expand $(x^2 + 2)^2 = x^4 + 4x^2 + 4$ - Expand $(Bx + C)(x - 1)(x^2 + 2)$: First, $(x - 1)(x^2 + 2) = x^3 - x^2 + 2x - 2$ Then multiply by $(Bx + C)$: $$ (Bx + C)(x^3 - x^2 + 2x - 2) = Bx^4 - Bx^3 + 2Bx^2 - 2Bx + Cx^3 - Cx^2 + 2Cx - 2C $$ - Expand $(Dx + E)(x - 1) = Dx^2 - Dx + Ex - E$ 6. **Combine all terms:** $$x^4 - x^3 + 2x^2 - x + 2 = A(x^4 + 4x^2 + 4) + Bx^4 - Bx^3 + 2Bx^2 - 2Bx + Cx^3 - Cx^2 + 2Cx - 2C + Dx^2 - Dx + Ex - E$$ 7. **Group like powers of $x$:** $$x^4: A + B$$ $$x^3: -B + C$$ $$x^2: 4A + 2B - C + D$$ $$x^1: -2B + 2C - D + E$$ $$x^0: 4A - 2C - E$$ 8. **Set up system of equations by matching coefficients:** $$\begin{cases} A + B = 1 \\ -B + C = -1 \\ 4A + 2B - C + D = 2 \\ -2B + 2C - D + E = -1 \\ 4A - 2C - E = 2 \end{cases}$$ 9. **Solve the system:** From equation 1: $B = 1 - A$ From equation 2: $C = -1 + B = -1 + 1 - A = -A$ Substitute $B$ and $C$ into equation 3: $$4A + 2(1 - A) - (-A) + D = 2 \Rightarrow 4A + 2 - 2A + A + D = 2 \Rightarrow (4A - 2A + A) + D + 2 = 2 \Rightarrow 3A + D + 2 = 2$$ $$\Rightarrow D = 2 - 2 - 3A = -3A$$ Substitute $B$, $C$, $D$ into equation 4: $$-2(1 - A) + 2(-A) - (-3A) + E = -1$$ $$-2 + 2A - 2A + 3A + E = -1$$ $$(-2) + 3A + E = -1$$ $$E = -1 + 2 - 3A = 1 - 3A$$ Substitute $A$, $C$, $E$ into equation 5: $$4A - 2(-A) - (1 - 3A) = 2$$ $$4A + 2A - 1 + 3A = 2$$ $$9A - 1 = 2$$ $$9A = 3$$ $$A = \frac{1}{3}$$ Then: $$B = 1 - \frac{1}{3} = \frac{2}{3}$$ $$C = -\frac{1}{3}$$ $$D = -3 \times \frac{1}{3} = -1$$ $$E = 1 - 3 \times \frac{1}{3} = 0$$ 10. **Rewrite partial fractions:** $$\frac{x^4 - x^3 + 2x^2 - x + 2}{(x - 1)(x^2 + 2)^2} = \frac{\frac{1}{3}}{x - 1} + \frac{\frac{2}{3}x - \frac{1}{3}}{x^2 + 2} + \frac{-x}{(x^2 + 2)^2}$$ 11. **Integrate each term separately:** $$\int \frac{1/3}{x - 1} dx = \frac{1}{3} \ln|x - 1| + C_1$$ $$\int \frac{\frac{2}{3}x - \frac{1}{3}}{x^2 + 2} dx = \frac{2}{3} \int \frac{x}{x^2 + 2} dx - \frac{1}{3} \int \frac{1}{x^2 + 2} dx$$ - For $\int \frac{x}{x^2 + 2} dx$, use substitution $u = x^2 + 2$, $du = 2x dx$: $$\int \frac{x}{x^2 + 2} dx = \frac{1}{2} \int \frac{du}{u} = \frac{1}{2} \ln|x^2 + 2| + C_2$$ - For $\int \frac{1}{x^2 + 2} dx$, recall: $$\int \frac{1}{x^2 + a^2} dx = \frac{1}{a} \arctan \frac{x}{a} + C$$ Here, $a = \sqrt{2}$, so: $$\int \frac{1}{x^2 + 2} dx = \frac{1}{\sqrt{2}} \arctan \frac{x}{\sqrt{2}} + C_3$$ Therefore: $$\int \frac{\frac{2}{3}x - \frac{1}{3}}{x^2 + 2} dx = \frac{2}{3} \times \frac{1}{2} \ln|x^2 + 2| - \frac{1}{3} \times \frac{1}{\sqrt{2}} \arctan \frac{x}{\sqrt{2}} + C = \frac{1}{3} \ln|x^2 + 2| - \frac{1}{3 \sqrt{2}} \arctan \frac{x}{\sqrt{2}} + C$$ 12. **Integrate the last term:** $$\int \frac{-x}{(x^2 + 2)^2} dx = - \int \frac{x}{(x^2 + 2)^2} dx$$ Use substitution $u = x^2 + 2$, $du = 2x dx$, so $x dx = \frac{du}{2}$: $$- \int \frac{x}{(x^2 + 2)^2} dx = - \int \frac{1}{u^2} \times \frac{du}{2} = - \frac{1}{2} \int u^{-2} du = - \frac{1}{2} \left(- \frac{1}{u} \right) + C = \frac{1}{2(x^2 + 2)} + C$$ 13. **Combine all results:** $$\int \frac{x^4 - x^3 + 2x^2 - x + 2}{(x - 1)(x^2 + 2)^2} dx = \frac{1}{3} \ln|x - 1| + \frac{1}{3} \ln|x^2 + 2| - \frac{1}{3 \sqrt{2}} \arctan \frac{x}{\sqrt{2}} + \frac{1}{2(x^2 + 2)} + C$$