Question

(a) Find \[ \int \frac{x}{(x - 1)(x^2 + 4)} \, dx \]

Hint:

When using partial fractions, match the degrees of the numerator and denominator. The result will simplify into integrable terms.

Answer 1

We use the method of partial fractions to decompose the integrand. We need to express: \[ \frac{x}{(x - 1)(x^2 + 4)} = \frac{A}{x - 1} + \frac{Bx + C}{x^2 + 4} \] Multiply both sides by \( (x - 1)(x^2 + 4) \): \[ x = A(x^2 + 4) + (Bx + C)(x - 1) \] Now expand both sides: \[ x = A(x^2 + 4) + Bx(x - 1) + C(x - 1) \] Simplify and collect like terms: \[ x = A x^2 + 4A + Bx^2 - Bx + Cx - C \] \[ x = (A + B) x^2 + (-B + C) x + (4A - C) \] Equating the coefficients of like powers of \( x \) on both sides: 1. \( A + B = 0 \) 2. \( -B + C = 1 \) 3. \( 4A - C = 0 \) Solve these equations to find \( A \), \( B \), and \( C \). After solving: \[ A = \frac{1}{5}, \quad B = -\frac{1}{5}, \quad C = \frac{1}{5} \] Thus, the integral becomes: \[ \int \frac{1}{5(x - 1)} + \int \frac{-\frac{1}{5}x + \frac{1}{5}}{x^2 + 4} \, dx \] Now, integrate term by term: \[ \frac{1}{5} \ln |x - 1| - \frac{1}{5} \int \frac{x}{x^2 + 4} \, dx + \frac{1}{5} \int \frac{1}{x^2 + 4} \, dx \] Use the substitution \( u = x^2 + 4 \) for the second integral and recognize that the third is a standard integral: \[ \frac{1}{5} \ln |x - 1| - \frac{1}{10} \ln(x^2 + 4) + \frac{1}{5} \tan^{-1}\left(\frac{x}{2}\right) + C \] Thus, the final result is: \[ \frac{1}{5} \ln |x - 1| - \frac{1}{10} \ln(x^2 + 4) + \frac{1}{5} \tan^{-1}\left(\frac{x}{2}\right) + C \]