1. **Problem:** Evaluate the integral $$\int \frac{dx}{x^2(x+1)}$$ **Step 1:** Use partial fraction decomposition: $$\frac{1}{x^2(x+1)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x+1}$$ **Step 2:** Multiply both sides by $x^2(x+1)$: $$1 = A x (x+1) + B (x+1) + C x^2$$ **Step 3:** Expand and group terms: $$1 = A x^2 + A x + B x + B + C x^2 = (A + C) x^2 + (A + B) x + B$$ **Step 4:** Equate coefficients: - For $x^2$: $A + C = 0$ - For $x$: $A + B = 0$ - Constant: $B = 1$ **Step 5:** Solve system: - $B = 1$ - $A + 1 = 0 \Rightarrow A = -1$ - $-1 + C = 0 \Rightarrow C = 1$ **Step 6:** Rewrite integral: $$\int \left(-\frac{1}{x} + \frac{1}{x^2} + \frac{1}{x+1}\right) dx = \int -\frac{1}{x} dx + \int \frac{1}{x^2} dx + \int \frac{1}{x+1} dx$$ **Step 7:** Integrate each term: $$-\ln|x| - \frac{1}{x} + \ln|x+1| + C$$ --- 2. **Problem:** Evaluate the definite integral $$\int_0^1 \frac{4x^2}{x^4 - 1} dx$$ **Step 1:** Factor denominator: $$x^4 - 1 = (x^2 - 1)(x^2 + 1) = (x-1)(x+1)(x^2 + 1)$$ **Step 2:** Use partial fractions: $$\frac{4x^2}{(x-1)(x+1)(x^2+1)} = \frac{A}{x-1} + \frac{B}{x+1} + \frac{Cx + D}{x^2 + 1}$$ **Step 3:** Multiply both sides by denominator and equate coefficients: $$4x^2 = A(x+1)(x^2+1) + B(x-1)(x^2+1) + (Cx + D)(x^2 - 1)$$ **Step 4:** Expand and group terms, then solve for $A,B,C,D$: - Coefficients yield: $A=1$, $B=-1$, $C=0$, $D=-2$ **Step 5:** Rewrite integral: $$\int_0^1 \left( \frac{1}{x-1} - \frac{1}{x+1} - \frac{2}{x^2+1} \right) dx$$ **Step 6:** Integrate each term: $$\left[ \ln|x-1| - \ln|x+1| - 2 \arctan x \right]_0^1$$ **Step 7:** Evaluate at bounds: - At $x=1$: $\ln|0|$ diverges, but integral is improper; consider limit from left. - At $x=0$: $\ln| -1| = \ln 1 = 0$, $\arctan 0 = 0$ **Step 8:** Evaluate limit as $x \to 1^-$: $$\lim_{x \to 1^-} (\ln|x-1| - \ln|x+1| - 2 \arctan x) = -\infty$$ **Step 9:** Integral diverges to $-\infty$; thus, the integral does not converge. --- 3. **Problem:** Evaluate $$\int \frac{dx}{x^3 + 1}$$ **Step 1:** Factor denominator: $$x^3 + 1 = (x+1)(x^2 - x + 1)$$ **Step 2:** Partial fractions: $$\frac{1}{(x+1)(x^2 - x + 1)} = \frac{A}{x+1} + \frac{Bx + C}{x^2 - x + 1}$$ **Step 3:** Multiply both sides by denominator: $$1 = A(x^2 - x + 1) + (Bx + C)(x+1)$$ **Step 4:** Expand and group terms: $$1 = A x^2 - A x + A + B x^2 + B x + C x + C$$ Group by powers: $$1 = (A + B) x^2 + (-A + B + C) x + (A + C)$$ **Step 5:** Equate coefficients: - $x^2$: $A + B = 0$ - $x$: $-A + B + C = 0$ - Constant: $A + C = 1$ **Step 6:** Solve system: - From $A + B = 0$, $B = -A$ - From $A + C = 1$, $C = 1 - A$ - Substitute into $-A + B + C = 0$: $$-A + (-A) + (1 - A) = 0 \Rightarrow -3A + 1 = 0 \Rightarrow A = \frac{1}{3}$$ Then: $$B = -\frac{1}{3}, \quad C = 1 - \frac{1}{3} = \frac{2}{3}$$ **Step 7:** Rewrite integral: $$\int \left( \frac{1/3}{x+1} + \frac{-\frac{1}{3} x + \frac{2}{3}}{x^2 - x + 1} \right) dx$$ **Step 8:** Integrate each term: - $$\int \frac{1/3}{x+1} dx = \frac{1}{3} \ln|x+1|$$ - For $$\int \frac{-\frac{1}{3} x + \frac{2}{3}}{x^2 - x + 1} dx$$, split: $$= -\frac{1}{3} \int \frac{x}{x^2 - x + 1} dx + \frac{2}{3} \int \frac{1}{x^2 - x + 1} dx$$ **Step 9:** Use substitution for the first integral: Let $$u = x^2 - x + 1, \quad du = (2x - 1) dx$$ Rewrite numerator: $$x = \frac{1}{2}(2x - 1) + \frac{1}{2}$$ So: $$\int \frac{x}{u} dx = \frac{1}{2} \int \frac{2x - 1}{u} dx + \frac{1}{2} \int \frac{1}{u} dx$$ The first part: $$\frac{1}{2} \int \frac{du}{u} = \frac{1}{2} \ln|u|$$ The second part is: $$\frac{1}{2} \int \frac{1}{x^2 - x + 1} dx$$ **Step 10:** The integral $$\int \frac{1}{x^2 - x + 1} dx$$ can be solved by completing the square: $$x^2 - x + 1 = \left(x - \frac{1}{2}\right)^2 + \frac{3}{4}$$ So: $$\int \frac{1}{(x - \frac{1}{2})^2 + (\frac{\sqrt{3}}{2})^2} dx = \frac{2}{\sqrt{3}} \arctan \left( \frac{2x - 1}{\sqrt{3}} \right) + C$$ **Step 11:** Combine all parts: $$\int \frac{x}{x^2 - x + 1} dx = \frac{1}{2} \ln|x^2 - x + 1| + \frac{1}{2} \cdot \frac{2}{\sqrt{3}} \arctan \left( \frac{2x - 1}{\sqrt{3}} \right) + C = \frac{1}{2} \ln|x^2 - x + 1| + \frac{1}{\sqrt{3}} \arctan \left( \frac{2x - 1}{\sqrt{3}} \right) + C$$ **Step 12:** Substitute back into integral: $$-\frac{1}{3} \int \frac{x}{x^2 - x + 1} dx = -\frac{1}{3} \left( \frac{1}{2} \ln|x^2 - x + 1| + \frac{1}{\sqrt{3}} \arctan \left( \frac{2x - 1}{\sqrt{3}} \right) \right)$$ **Step 13:** The other integral: $$\frac{2}{3} \int \frac{1}{x^2 - x + 1} dx = \frac{2}{3} \cdot \frac{2}{\sqrt{3}} \arctan \left( \frac{2x - 1}{\sqrt{3}} \right) = \frac{4}{3 \sqrt{3}} \arctan \left( \frac{2x - 1}{\sqrt{3}} \right)$$ **Step 14:** Combine arctan terms: $$-\frac{1}{3 \sqrt{3}} + \frac{4}{3 \sqrt{3}} = \frac{3}{3 \sqrt{3}} = \frac{1}{\sqrt{3}}$$ **Step 15:** Final answer: $$\frac{1}{3} \ln|x+1| - \frac{1}{6} \ln|x^2 - x + 1| + \frac{1}{\sqrt{3}} \arctan \left( \frac{2x - 1}{\sqrt{3}} \right) + C$$ --- 4. **Problem:** Evaluate $$\int \frac{dx}{x^3 + 4x}$$ **Step 1:** Factor denominator: $$x^3 + 4x = x(x^2 + 4)$$ **Step 2:** Partial fractions: $$\frac{1}{x(x^2 + 4)} = \frac{A}{x} + \frac{Bx + C}{x^2 + 4}$$ **Step 3:** Multiply both sides by denominator: $$1 = A(x^2 + 4) + (Bx + C) x = A x^2 + 4A + B x^2 + C x$$ Group terms: $$1 = (A + B) x^2 + C x + 4A$$ **Step 4:** Equate coefficients: - $x^2$: $A + B = 0$ - $x$: $C = 0$ - Constant: $4A = 1 \Rightarrow A = \frac{1}{4}$ **Step 5:** Solve for $B$ and $C$: - $B = -A = -\frac{1}{4}$ - $C = 0$ **Step 6:** Rewrite integral: $$\int \left( \frac{1/4}{x} + \frac{-\frac{1}{4} x}{x^2 + 4} \right) dx = \frac{1}{4} \int \frac{1}{x} dx - \frac{1}{4} \int \frac{x}{x^2 + 4} dx$$ **Step 7:** Integrate each term: - $$\frac{1}{4} \ln|x|$$ - For $$\int \frac{x}{x^2 + 4} dx$$, use substitution: Let $$u = x^2 + 4, du = 2x dx \Rightarrow x dx = \frac{du}{2}$$ So: $$\int \frac{x}{x^2 + 4} dx = \frac{1}{2} \int \frac{1}{u} du = \frac{1}{2} \ln|u| + C = \frac{1}{2} \ln|x^2 + 4| + C$$ **Step 8:** Substitute back: $$-\frac{1}{4} \cdot \frac{1}{2} \ln|x^2 + 4| = -\frac{1}{8} \ln|x^2 + 4|$$ **Step 9:** Final answer: $$\frac{1}{4} \ln|x| - \frac{1}{8} \ln|x^2 + 4| + C$$