Question
(CBSE 2023 style.) Evaluate ∫(x2+1)(x2+4)x2dx using partial fractions.
Solution — Step by Step
We want to decompose:
(x2+1)(x2+4)x2=x2+1A+x2+4B
Both denominators are quadratic and irreducible (no real roots), so the numerators are constants — let them be A and B. (If they were linear, we’d use Ax+B form, but the original numerator has only x2, so we can guess constants will work.)
Multiply both sides by (x2+1)(x2+4):
x2=A(x2+4)+B(x2+1)
Equating coefficients of x2: A+B=1. Equating constants: 4A+B=0.
Subtracting: 3A=−1⇒A=−1/3, B=4/3.
∫(x2+1)(x2+4)x2dx=−31∫x2+1dx+34∫x2+4dx
=−31tan−1x+34⋅21tan−12x+C
=−31tan−1x+32tan−12x+C
Why This Works
Partial fractions split a complicated rational function into a sum of simpler ones whose integrals are standard. For irreducible quadratic factors x2+a2, the standard integral ∫x2+a2dx=a1tan−1(x/a)+C takes care of each piece.
The key trick when both denominators are even-power polynomials: assume constants in the numerators of partial fractions. If the original numerator is purely even-degree, this assumption holds. If it has odd-degree terms, you’d need linear numerators (Ax+B).
Quick sanity check: differentiate your answer mentally. If the derivative looks like the original integrand, you are correct. Catching errors at this step saves marks.
Alternative Method
Add-subtract trick: write
(x2+1)(x2+4)x2=(x2+1)(x2+4)x2+4−4=x2+11−(x2+1)(x2+4)4
Or: write x2=(x2+1)−1=(x2+4)−4, then split. Either way leads to the same partial fraction structure, just with different intermediate algebra.
Students forget to factor before integrating. The denominators x2+1 and x2+4 have no real roots, so they don’t factor over reals — but the partial fraction works fine because we keep them as quadratics. Don’t try to factor them as (x±i)(x∓i) — that complicates things.
Final answer: −31tan−1x+32tan−1(x/2)+C.