∫ (x²dx)/((x²+a²)(x²+b²))

NCERT Exemplar · Q36

\int \dfrac{x^2\,dx}{(x^2+a^2)(x^2+b^2)}

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Concept understanding — Partial Fraction Decomposition

Partial Fraction Decomposition

Adding $\frac{2}{x-1}+\frac{3}{x+2}$ over a common denominator gives $\frac{5x+1}{x^2+x-2}$ . Partial fraction decomposition reverses this — it breaks one complicated rational function back into a sum of simple pieces. Those pieces are far easier to integrate: $\int \frac{2}{x-1}\,dx = 2\log|x-1|$ is immediate, while the combined fraction is not.

When it applies

You need a proper rational function, $\deg P < \deg Q$ . If the numerator's degree is equal or higher, first do polynomial long division. The denominator $Q(x)$ must be factored into linear and/or irreducible quadratic factors.

The standard forms

For $\frac{P(x)}{Q(x)}$ with $Q$ fully factored, each factor contributes a term:

Distinct linear $(ax+b)$ $\to$ $\dfrac{A}{ax+b}$ .
Repeated linear $(ax+b)^n$ $\to$ $\dfrac{A_1}{ax+b}+\dfrac{A_2}{(ax+b)^2}+\cdots+\dfrac{A_n}{(ax+b)^n}$ .
Irreducible quadratic $(ax^2+bx+c)$ $\to$ $\dfrac{Ax+B}{ax^2+bx+c}$ — a linear numerator, not just a constant.

The unknown constants are found from the resulting equations; the decomposition is unique, which is what lets us solve for them systematically.

Worked example

Decompose $\frac{3x+5}{x^2-3x+2}$ . Factor the denominator: $(x-1)(x-2)$ . Set

\frac{3x+5}{(x-1)(x-2)} = \frac{A}{x-1}+\frac{B}{x-2}.

Clear denominators: $3x+5 = A(x-2)+B(x-1)$ . Substituting the roots, $x=1$ gives $8=-A$ so $A=-8$ , and $x=2$ gives $11=B$ . Hence

\frac{3x+5}{x^2-3x+2} = \frac{-8}{x-1}+\frac{11}{x-2}.

Tip

With distinct linear factors, substitute each factor's root to knock out all but one term — much faster than equating coefficients. …

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We decompose the integrand into simpler fractions using the method of partial fractions, exploiting the fact that the denominator factors as a product of quadratics. The integral evaluates to $\frac{1}{b^2 - a^2} \left( b \tan^{-1}\frac{x}{b} - a \tan^{-1}\frac{x}{a} \right) + C$ .

The key insight here is that the denominator is a product of two irreducible quadratics: $(x^2 + a^2)$ and $(x^2 + b^2)$ . When we have a rational function where the numerator is of lower degree than the denominator, partial fraction decomposition lets us break it into a sum of simpler fractions — each with a single quadratic denominator. This turns a messy integral into two standard arctangent integrals.

The trick is to find constants $A$ and $B$ such that:

\frac{x^2}{(x^2 + a^2)(x^2 + b^2)} = \frac{A}{x^2 + a^2} + \frac{B}{x^2 + b^2}

Why does this work? Because the numerator $x^2$ is of degree 2, and each denominator is degree 2, so the partial fractions have constant numerators (not linear ones). If the numerator were degree 1 or higher, we'd need linear numerators like $Cx + D$ , but here it's just constants.

Let's find $A$ and $B$ .

Set up the equation. Multiply both sides by the common denominator $(x^2 + a^2)(x^2 + b^2)$ :

x^2 = A(x^2 + b^2) + B(x^2 + a^2)

Expand and collect like terms:

x^2 = A x^2 + A b^2 + B x^2 + B a^2

x^2 = (A + B)x^2 + (A b^2 + B a^2)

Equate coefficients. For this to hold for all $x$ , the coefficients of $x^2$ and the constant term must match on both sides:
- Coefficient of $x^2$ : $A + B = 1$
- Constant term: $A b^2 + B a^2 = 0$
Solve the system. From the second equation: $A b^2 = -B a^2$ , so $A = -\frac{B a^2}{b^2}$ . Substitute into $A + B = 1$ :

-\frac{B a^2}{b^2} + B = 1

B\left(1 - \frac{a^2}{b^2}\right) = 1

B\left(\frac{b^2 - a^2}{b^2}\right) = 1

B = \frac{b^2}{b^2 - a^2}

Then $A = 1 - B = 1 - \frac{b^2}{b^2 - a^2} = \frac{b^2 - a^2 - b^2}{b^2 - a^2} = \frac{-a^2}{b^2 - a^2}$ .

So we have:

A = \frac{-a^2}{b^2 - a^2}, \quad B = \frac{b^2}{b^2 - a^2}

Tip

Notice the symmetry: $A$ and $B$ are just swapped roles of $a$ and $b$ , with a sign difference. This is a good sanity check — if you swap $a$ and $b$ , the original integrand stays the same, and the decomposition should reflect that.

Rewrite the integral. Substituting back: …

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Q. $\int \dfrac{x^2\,dx}{(x^2+a^2)(x^2+b^2)}$

Concept understanding — Partial Fraction Decomposition

Partial Fraction Decomposition

When it applies

The standard forms

Worked example

Method: Splitting a product of two distinct quadratics

Steps

Common Mistakes

Q.∫x2 dx(x2+a2)(x2+b2)\int \dfrac{x^2\,dx}{(x^2+a^2)(x^2+b^2)}∫(x2+a2)(x2+b2)x2dx​

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Q. $\int \dfrac{x^2\,dx}{(x^2+a^2)(x^2+b^2)}$

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