Partial Fraction Integration 919C6A

1. **State the problem:** We need to evaluate the integral $$\int \frac{2x + 1}{(x - 1)(x + 3)} \, dx.$$ by decomposing the rational function into partial fractions first. 2. **Partial fraction decomposition formula:** For a rational function $$\frac{P(x)}{(x - a)(x - b)}$$ where the denominator factors are linear and distinct, we write: $$\frac{P(x)}{(x - a)(x - b)} = \frac{A}{x - a} + \frac{B}{x - b}$$ where $$A$$ and $$B$$ are constants to be determined. 3. **Apply to our problem:** $$\frac{2x + 1}{(x - 1)(x + 3)} = \frac{A}{x - 1} + \frac{B}{x + 3}$$ Multiply both sides by $$ (x - 1)(x + 3) $$: $$2x + 1 = A(x + 3) + B(x - 1)$$ 4. **Expand and group terms:** $$2x + 1 = A x + 3A + B x - B$$ $$2x + 1 = (A + B) x + (3A - B)$$ 5. **Equate coefficients:** For $$x$$ terms: $$2 = A + B$$ For constant terms: $$1 = 3A - B$$ 6. **Solve the system:** From $$2 = A + B$$, we get $$B = 2 - A$$. Substitute into $$1 = 3A - B$$: $$1 = 3A - (2 - A) = 3A - 2 + A = 4A - 2$$ Add 2 to both sides: $$3 = 4A$$ Divide both sides by 4: $$A = \frac{3}{4}$$ 7. **Find $$B$$:** $$B = 2 - A = 2 - \frac{3}{4} = \frac{8}{4} - \frac{3}{4} = \frac{5}{4}$$ 8. **Rewrite the integral:** $$\int \frac{2x + 1}{(x - 1)(x + 3)} \, dx = \int \frac{3/4}{x - 1} \, dx + \int \frac{5/4}{x + 3} \, dx$$ 9. **Integrate each term:** $$= \frac{3}{4} \int \frac{1}{x - 1} \, dx + \frac{5}{4} \int \frac{1}{x + 3} \, dx$$ 10. **Recall integral of $$\frac{1}{x - a}$$ is $$\ln|x - a| + C$$:** $$= \frac{3}{4} \ln|x - 1| + \frac{5}{4} \ln|x + 3| + C$$ **Final answer:** $$\int \frac{2x + 1}{(x - 1)(x + 3)} \, dx = \frac{3}{4} \ln|x - 1| + \frac{5}{4} \ln|x + 3| + C$$