calculate the integral question using partial fraction

Question

Find ∫(4x − 1) /(x3 − x2 + 2) dx.

Answer 1

(4x − 1) /(x³ − x² + 2) = (4x − 1) /(x + 1)(x² – 2x + 2) = A/(x + 1) + (Bx + C)/(x² – 2x + 2)

A(x² – 2x + 2) + (Bx + C)(x + 1) = (A + B)x² + (B + C – 2A)x + (2A + C) = 4x – 1

A + B = 0 B + C – 2A = 3B + C = 4 2A + C = –2B + C = –1 ⇒ 2B – C = 1

⇒ (3B + C) + (2B – C) = (4 + 1) ⇒ 5B = 5 ⇒ B = 1

A + B = 0 ⇒ A = –1

3B + C = 4 ⇒ C = 1

(4x − 1) /(x + 1)(x² – 2x + 2) = –1/(x + 1) + (x + 1)/(x² – 2x + 2) = –1/(x + 1) + (x + 1)/((x – 1)² + 1)

∫[(4x − 1) /(x³ − x² + 2)]dx = –∫dx/(x + 1) + ∫[(x + 1)/((x – 1)² + 1)]dx = –ln|x+1| + ∫[(x + 1)/((x – 1)² + 1)]dx

∫[(x + 1)/((x – 1)² + 1)]dx Let u = x – 1 then ∫[(x + 1)/((x – 1)² + 1)]dx = ∫[(u + 2)/(u² + 1)]du

= ∫[u/(u² + 1)]du + 2∫[1/(u² + 1)]du = ½ln(u² + 1) + 2 tan⁻¹(u)

Substitute back: u = x – 1 and u² + 1 = (x – 1)² + 1 = x² – 2x + 2

∫[(x + 1)/((x – 1)² + 1)]dx = ½ln(x² – 2x + 2) + 2 tan⁻¹(x – 1) + C

Answer

∫[(4x − 1) /(x³ − x² + 2)]dx = –ln|x+1| + ½ln(x² – 2x + 2) + 2 tan⁻¹(x – 1) + C