∫ (x²dx)/(x⁴-x²-12)

NCERT Exemplar · Q35

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

Jammu Kashmir JkboseLong· 3mImportance★★★★★

90% · 337/373 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

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. …

Start your 14-day free trial to unlock the full solution →

Factor the denominator as $(x^2-4)(x^2+3)$ and split as a partial fraction in $x^2$ . The result is $\frac{1}{7}\log\left|\frac{x-2}{x+2}\right| + \frac{\sqrt{3}}{7}\tan^{-1}\frac{x}{\sqrt{3}} + C$ .

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

x^4 - x^2 - 12 = (x^2-4)(x^2+3) = (x-2)(x+2)(x^2+3).

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

\frac{u}{(u-4)(u+3)} = \frac{A}{u-4}+\frac{B}{u+3}.

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

\frac{x^2}{x^4-x^2-12} = \frac{4/7}{x^2-4} + \frac{3/7}{x^2+3}.

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

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

Start your 14-day free trial to unlock the full solution →

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

JKBOSE Class 12 Annual Regular Examination 2022Set SZ6 marks
$\dfrac{2x - 3}{(x^2 - 1)(2x + 3)}$
›Reveal solutionSolution
This question offers a choice. Alternative 1 integrates a rational function by partial fractions. Alternative 2 evaluates a definite integral of an absolute-value function by splitting at the point where the expression inside changes sign.

Alternative 1 — integrate $\dfrac{2x-3}{(x^2-1)(2x+3)}$ :

Factor $x^2-1=(x-1)(x+1)$ , so the denominator is $(x-1)(x+1)(2x+3)$ . Write the partial fraction decomposition:

$\frac{2x-3}{(x-1)(x+1)(2x+3)}=\frac{A}{x-1}+\frac{B}{x+1}+\frac{C}{2x+3}$

Multiplying through:

$2x-3=A(x+1)(2x+3)+B(x-1)(2x+3)+C(x-1)(x+1)$

Plug in convenient values of $x$ :

$x=1$ : $2-3=-1=A(2)(5)=10A\Rightarrow A=-\dfrac1{10}$

$x=-1$ : $-2-3=-5=B(-2)(1)=-2B\Rightarrow B=\dfrac52$

$x=-\dfrac32$ : $-3-3=-6=C\left(-\dfrac52\right)\left(-\dfrac12\right)=C\cdot\dfrac54\Rightarrow C=-\dfrac{24}{5}$

So:

$\int\frac{2x-3}{(x^2-1)(2x+3)}dx=\int\left[\frac{-1/10}{x-1}+\frac{5/2}{x+1}+\frac{-24/5}{2x+3}\right]dx$

$=-\frac{1}{10}\ln|x-1|+\frac{5}{2}\ln|x+1|-\frac{24}{5}\cdot\frac12\ln|2x+3|+C$

$=-\frac{1}{10}\ln|x-1|+\frac{5}{2}\ln|x+1|-\frac{12}{5}\ln|2x+3|+C$

Alternative 2 — evaluate $\displaystyle\int_{-5}^{5}|x+2|\,dx$ :

The expression $x+2$ changes sign at $x=-2$ , which lies inside $[-5,5]$ . So split the integral there: …
Start your 14-day free trial to see the full solution →
JKBOSE Class 12 Annual Regular Examination 2020Set SZ6 marks
$\displaystyle\int \dfrac{\cos x\, dx}{(1-\sin x)(2-\sin x)}$
›Reveal solutionSolution

Substitute $t=\sin x$ , then split into partial fractions.

Main part. Let $t=\sin x$ , so $dt=\cos x\,dx$ . The integral becomes $\displaystyle\int \dfrac{dt}{(1-t)(2-t)}$ .

Partial fractions: $\dfrac{1}{(1-t)(2-t)} = \dfrac{A}{1-t}+\dfrac{B}{2-t}$ . Solving, $A=1,\ B=-1$ .

$\displaystyle\int\left[\dfrac{1}{1-t}-\dfrac{1}{2-t}\right]dt = -\ln|1-t| + \ln|2-t| + C = \ln\left|\dfrac{2-t}{1-t}\right|+C$

Substituting back $t=\sin x$ : $\ln\left|\dfrac{2-\sin x}{1-\sin x}\right| + C$ .

OR (alternative part). $\displaystyle\int_0^2(x^2+1)dx$ as a limit of a sum: with $h=\dfrac{2-0}{n}=\dfrac2n$ ,

$\displaystyle\int_0^2 f(x)dx = \lim_{n\to\infty} h\sum_{r=0}^{n-1} f(rh)$ , $f(x)=x^2+1$ .

$S_n = h\sum_{r=0}^{n-1}\left[(rh)^2+1\right] = h^3\sum r^2 + hn = h^3\cdot\dfrac{(n-1)n(2n-1)}{6} + hn$

…

Start your 14-day free trial to see the full solution →

←Previous questionQ34 — Evaluate: ∫₀¹ dfracdx(1+x²)√(1-x²) (Hint: let x=sinθ)→Next questionQ36 — ∫ (x²dx)/((x²+a²)(x²+b²))

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

Concept understanding — Partial Fraction Decomposition

Partial Fraction Decomposition

When it applies

The standard forms

Worked example

Method: Partial fractions on a biquadratic denominator

Steps

Common Mistakes

Q.∫x2 dxx4−x2−12\int \dfrac{x^2\,dx}{x^4-x^2-12}∫x4−x2−12x2dx​

🎓Unlock everything free for 14 days

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

Unlock everything free for 14 days