Use partial fractions to evaluate the integral

Question

Use partial fractions to evaluate the integral

∫ (4x^3 + 2x^2 + 3x + 2) / (x^2(x^2 + 1)) dx

Answer 1

(4x³ + 2x² + 3x + 2) / [x²(x² + 1)] = (Ax + B) / x² + (Cx + D) / (x² + 1)

(Ax + B)(x² + 1) + (Cx + D)x² = 4x³ + 2x² + 3x + 2

Ax³ + Bx² + Ax + B + Cx³ +Dx² = (A + C)x³ + (B + D)x² + Ax + B = 4x³ + 2x² + 3x + 2

A + C = 4     B + D = 2     A = 3     B = 2

So C = 1     D = 0

(4x³ + 2x² + 3x + 2) / [x²(x² + 1)] = (Ax + B) / x² + (Cx + D) / (x² + 1) = (3x + 2) / x² + x / (x² + 1)

∫((4x³ + 2x² + 3x + 2) / [x²(x² + 1)]) dx = ∫[(3x + 2) / x²] dx + ∫[x / (x² + 1)] dx

= ∫(3 / x)dx + ∫(2 / x²)dx + ∫[x / (x² + 1)] dx

= 3 ln|x| –2/x + ∫[x / (x² + 1)] dx

∫[x / (x² + 1)] dx let u = x² + 1, du = 2xdx, dx = ½du

∫[x / (x² + 1)] dx= ½∫du/u = ½ln|u| = ½ln(x² + 1) we can get rid of the absolute value because x² + 1 ≥ 1

Answer

∫((4x³ + 2x² + 3x + 2) / [x²(x² + 1)]) dx = 3 ln|x| –2/x + ½ln(x² + 1) + C

Answer 2

desmos.com/calculator/z9w8mforfg