Q.Find $\int \dfrac{x^2}{(x^2+1)(x^2+4)}\, dx$

Concept: Partial Fraction Decomposition — splitting the integrand into simpler fractions whose denominators are the irreducible quadratic factors.

First, note the degree of numerator and denominator are equal, so we rewrite:

Multiply through and compare numerators:

Equating coefficients gives $A+B = 1$ and $4A+B = 0$. Solving: $A = -\frac13$, $B = \frac43$.

Thus the integral becomes: …

Why partial fractions work here

When you see a rational function where the denominator is already factored into quadratics, and the numerator is of lower degree than the denominator, partial fraction decomposition is the natural tool. The idea: break a complicated fraction into a sum of simpler ones that we know how to integrate.

Here, the denominator is $(x^2+1)(x^2+4)$. Both factors are irreducible quadratics (they have no real roots). So the decomposition will take the form:

Why linear numerators? Because for an irreducible quadratic denominator, the numerator in the partial fraction must be one degree less — that is, linear.

Step-by-step solution

1. Set up the decomposition

Multiply both sides by $(x^2+1)(x^2+4)$:

2. Expand and collect like terms

3. Equate coefficients

Since the left side is $0x^3 + 1x^2 + 0x + 0$, we get:

From $A + C = 0$ and $4A + C = 0$, subtract the first from the second: $(4A + C) - (A + C) = 0 - 0 \implies 3A = 0 \implies A = 0$. Then $C = 0$.

From $B + D = 1$ and $4B + D = 0$, subtract: $(4B + D) - (B + D) = 0 - 1 \implies 3B = -1 \implies B = -\frac{1}{3}$. Then $D = 1 - B = 1 + \frac{1}{3} = \frac{4}{3}$.

4. Write the decomposed form

Or more neatly:

5. Integrate term by term

$\int \frac{x^2}{(x^2+1)(x^2+4)} \, dx = -\frac{1}{3} \int \frac{dx}{x^2+1} + \frac{4}{3} \int \frac{dx}{x^2+4}$ …

Method: Partial Fractions via a Hidden Substitution (Even-Power Denominators)

Use this when the denominator is a product of irreducible quadratics in $x^2$: treating $x^2$ as a single variable turns it into an ordinary linear partial-fraction problem.

Steps

Step 1: Substitute $y=x^2$ to decompose.

For $\dfrac{x^2}{(x^2+1)(x^2+4)}$, write $\dfrac{y}{(y+1)(y+4)}=\dfrac{A}{y+1}+\dfrac{B}{y+4}$ and solve for $A,B$.

Step 2: Restore $x^2$ in the decomposition. …

Common Mistakes

Mistake 1: Trying a linear numerator $Bx+C$ over $x^2+1$ unnecessarily.

Why it's wrong: since the numerator involves only $x^2$, the decomposition needs only constant numerators over each quadratic. Correct approach: use $\frac{A}{x^2+1}+\frac{B}{x^2+4}$.

Mistake 2: Integrating in the substituted variable $y=x^2$. …