Integrals: Indefinite Integration by Partial Fractions

Let: $$\frac{x}{(x - 1)(x^2 + 1)} = \frac{A}{x - 1} + \frac{Bx + C}{x^2 + 1}$$ Multiplying both sides by $(x - 1)(x^2 + 1)$: $$x = A(x^2 + 1) + (Bx + C)(x - 1)$$

Substitute $x = 1$: $$1 = A(1^2 + 1) + 0 \implies 2A = 1 \implies A = \frac{1}{2}$$ Equating coefficients of $x^2$: $$0 = A + B \implies B = -A = -\frac{1}{2}$$ Equating constant terms ($x = 0$): $$0 = A - C \implies C = A = \frac{1}{2}$$ Thus: $$\frac{x}{(x - 1)(x^2 + 1)} = \frac{1/2}{x - 1} + \frac{-\frac{1}{2}x + \frac{1}{2}}{x^2 + 1} = \frac{1}{2(x - 1)} - \frac{1}{4}\left(\frac{2x}{x^2 + 1}\right) + \frac{1}{2}\left(\frac{1}{x^2 + 1}\right)$$

$$\int \frac{x}{(x - 1)(x^2 + 1)} \, dx = \frac{1}{2}\int \frac{dx}{x - 1} - \frac{1}{4}\int \frac{2x}{x^2 + 1} \, dx + \frac{1}{2}\int \frac{dx}{x^2 + 1}$$ $$= \frac{1}{2} \ln|x - 1| - \frac{1}{4} \ln(x^2 + 1) + \frac{1}{2} \tan^{-1} x + C$$