Q.$\int \dfrac{x^2\,dx}{x^4-x^2-12}$

Concept: Partial Fraction Decomposition — we factor the denominator and split the rational function into simpler fractions.

First, factor the denominator:

The integrand is a proper rational function (degree 2 < degree 4). Write:

Multiply through and equate coefficients. Solving gives:

Thus the integral becomes:

Integrate term by term: …

1. Factor the denominator. Treating it as a quadratic in $x^2$,

2. Partial fractions in $x^2$. Let $u=x^2$:

Setting $u=4$: $4=7A\Rightarrow A=\tfrac47$. Setting $u=-3$: $-3=-7B\Rightarrow B=\tfrac37$. Hence

3. Integrate each term. Using $\int\frac{dx}{x^2-a^2}=\frac{1}{2a}\log\left|\frac{x-a}{x+a}\right|$ and $\int\frac{dx}{x^2+a^2}=\frac1a\tan^{-1}\frac{x}{a}$:

$\frac47\int\frac{dx}{x^2-4} = \frac47\cdot\frac14\log\left|\frac{x-2}{x+2}\right| = \frac17\log\left|\frac{x-2}{x+2}\right|,$ …

Method: Partial fractions on a biquadratic denominator

Use this whenever a rational integrand has a denominator that is a polynomial in $x^2$ only (a "biquadratic" such as $x^4+px^2+q$) and the numerator is also built from $x^2$.

Steps

Step 1: Substitute $u=x^2$ to factor the denominator.

Treat the denominator as a quadratic in $u$, factor it, then return to $x$. A factor $u-k$ becomes $x^2-k$: if $k>0$ it splits further into real linear factors $(x-\sqrt k)(x+\sqrt k)$; if $k<0$ it stays as an irreducible quadratic $x^2+|k|$.

Step 2: Decompose with the correct template. …

Common Mistakes

Mistake 1: Trying to break $x^4-x^2-12$ straight into linear factors.

Why it's wrong: it is a quadratic in $x^2$, and one of its factors ($x^2+3$) is irreducible over the reals. Correct approach: factor as $(x^2-4)(x^2+3)$; only $x^2-4=(x-2)(x+2)$ splits further.

Mistake 2: Forgetting the $\frac{1}{2a}$ in $\int\frac{dx}{x^2-a^2}$. …