Integrals: Step-by-Step Worked Examples (4)

easy • 2 min read
Tags Integrals

Question

Evaluate ∫x2+1(x2+2)(x2+3) dx\displaystyle\int \frac{x^2 + 1}{(x^2 + 2)(x^2 + 3)}\,dx using partial fractions. JEE Main 2024 standard, also a CBSE 6-mark question.

Solution — Step by Step

Both denominators are irreducible quadratics in x2x^2. Treat x2x^2 as a single variable for partial fractions.

x2+1(x2+2)(x2+3)=Ax2+2+Bx2+3\frac{x^2 + 1}{(x^2 + 2)(x^2 + 3)} = \frac{A}{x^2 + 2} + \frac{B}{x^2 + 3}

Multiply both sides by (x2+2)(x2+3)(x^2 + 2)(x^2 + 3):

x2+1=A(x2+3)+B(x2+2)x^2 + 1 = A(x^2 + 3) + B(x^2 + 2)

Set x2=−2x^2 = -2: −2+1=A(1)+0  ⟹  A=−1-2 + 1 = A(1) + 0 \implies A = -1.

Set x2=−3x^2 = -3: −3+1=0+B(−1)  ⟹  B=2-3 + 1 = 0 + B(-1) \implies B = 2.

∫−1x2+2 dx+∫2x2+3 dx\int \frac{-1}{x^2 + 2}\,dx + \int \frac{2}{x^2 + 3}\,dx

Use ∫dxx2+a2=1atan⁡−1xa+C\int \dfrac{dx}{x^2 + a^2} = \dfrac{1}{a}\tan^{-1}\dfrac{x}{a} + C:

−12tan⁡−1x2+23tan⁡−1x3+C-\frac{1}{\sqrt{2}}\tan^{-1}\frac{x}{\sqrt{2}} + \frac{2}{\sqrt{3}}\tan^{-1}\frac{x}{\sqrt{3}} + C

∫x2+1(x2+2)(x2+3) dx=−12tan⁡−1x2+23tan⁡−1x3+C\boxed{\int \frac{x^2 + 1}{(x^2 + 2)(x^2 + 3)}\,dx = -\frac{1}{\sqrt{2}}\tan^{-1}\frac{x}{\sqrt{2}} + \frac{2}{\sqrt{3}}\tan^{-1}\frac{x}{\sqrt{3}} + C}

Why This Works

Partial fractions break a complicated rational function into simple pieces, each of which has a standard antiderivative. The trick of treating x2x^2 as a single variable works because both denominators contain only x2x^2 (no linear xx terms).

The standard form ∫dxx2+a2=1atan⁡−1xa\int \dfrac{dx}{x^2 + a^2} = \dfrac{1}{a}\tan^{-1}\dfrac{x}{a} is one of the seven or eight integrals every JEE candidate must memorise. It comes up in 60% of partial fraction problems.

Alternative Method

Try the substitution method first: divide numerator and denominator by x2x^2:

1+1/x2(x+2/x)(x+3/x)\frac{1 + 1/x^2}{(x + 2/x)(x + 3/x)}

This sometimes simplifies cleanly with t=x−k/xt = x - k/x. Here it doesn’t help much, so partial fractions wins.

A frequent error: writing Ax+Bx2+2+Cx+Dx2+3\dfrac{Ax + B}{x^2 + 2} + \dfrac{Cx + D}{x^2 + 3} instead of just Ax2+2+Bx2+3\dfrac{A}{x^2 + 2} + \dfrac{B}{x^2 + 3}. The numerators here can be constants because the original numerator x2+1x^2 + 1 is also “constant in terms of x2x^2”. Adding AxAx and CxCx creates degrees of freedom you don’t need and complicates the algebra.

Once you find AA and BB, verify by adding the fractions back. If A(x2+3)+B(x2+2)(x2+2)(x2+3)\dfrac{A(x^2+3) + B(x^2+2)}{(x^2+2)(x^2+3)} matches the original, you’re correct. Takes 30 seconds and saves you a wrong answer.

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