Math · Calculus II · Worked example
Partial fractions with an irreducible quadratic
Find ∫(3x² + 2x + 5)/((x − 1)(x² + 4)) dx.
Set up the terms
x² + 4 has no real zeros, so it gets a linear numerator: 3x² + 2x + 5 = A(x² + 4) + (Bx + C)(x − 1).
Find A
x = 1 gives 10 = 5A.
Find B and C
Subtracting 2(x² + 4) from the numerator leaves x² + 2x − 3 = (x − 1)(x + 3), so Bx + C = x + 3.
Integrate each piece
Split (x + 3)/(x² + 4) into x/(x² + 4), which gives ½ ln(x² + 4) by substitution, and 3/(x² + 4), which gives (3/2) arctan(x/2).
Result
2 ln|x − 1| + ½ ln(x² + 4) + (3/2) arctan(x/2) + C.
Your turn
Find ∫1/(x² + 9) dx and ∫2x/(x² + 9) dx.
Show the answer and explanation
(1/3) arctan(x/3) + C and ln(x² + 9) + C.
∫dx/(x² + b²) = (1/b) arctan(x/b) with b = 3, and the second integral is a substitution with u = x² + 9.
Keep exploring
Derivative & antiderivative checks verifies the antiderivative on x > 1.
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