Integrate the function (x²)/(1-x⁶) – Class 12 Mathematics | ConceptFirst

Exercise 7.4 · Q6

\frac{x^2}{1-x^6}

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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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Substituting $u=x^3$ , $\displaystyle\int\frac{x^2}{1-x^6}\,dx=\frac16\log\left|\frac{1+x^3}{1-x^3}\right|+C$ .

Note $1-x^6=1-(x^3)^2$ and the numerator $x^2$ is a constant multiple of the derivative of $x^3$ , so substitute $u=x^3$ .

Substitution: $u=x^3\Rightarrow du=3x^2\,dx\Rightarrow x^2\,dx=\tfrac13\,du$ :

\int\frac{x^2}{1-x^6}\,dx=\frac13\int\frac{du}{1-u^2}.

Standard integral:

\int\frac{du}{1-u^2}=\frac12\log\left|\frac{1+u}{1-u}\right|+C,

so …

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

CBSE 2026Set A1 markMCQ
$\int\frac{dx}{x^2 - a^2} =$
(a) $\frac{1}{a}\tan^{-1}\frac{x}{a} + k$
(b) $\frac{1}{2a}\log\left|\frac{x-a}{x+a}\right| + k$
(c) $\frac{1}{2a}\log\left|\frac{a+x}{a-x}\right| + k$
(d) $\frac{1}{a}\log\left|\frac{x-a}{x+a}\right| + k$
›Reveal solutionSolution

Standard integral: $\int\tfrac{dx}{x^2-a^2}=\tfrac{1}{2a}\log\left|\tfrac{x-a}{x+a}\right|+k$ .

Using partial fractions, $\dfrac{1}{x^2-a^2}=\dfrac{1}{(x-a)(x+a)}=\dfrac{1}{2a}\left(\dfrac{1}{x-a}-\dfrac{1}{x+a}\right)$ .

…

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CBSE 2024Set D1 markMCQ
$\int \frac{dx}{x(x+2)} =$
(a) $\log\left|\frac{x}{x+2}\right| + c$
(b) $\frac{1}{2}\log\left|\frac{x}{x+2}\right| + c$
(c) $\log|x| + c$
(d) $\log|x+2| + c$
›Reveal solutionSolution

Partial fractions give $\frac{1}{2}\log\left|\frac{x}{x+2}\right|+c$ .

Write $\dfrac{1}{x(x+2)} = \dfrac{1}{2}\left(\dfrac{1}{x} - \dfrac{1}{x+2}\right)$ .

…

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CBSE 2022Set ANNUAL1 markMCQ
$\int \frac{dx}{x^2 - 1} =$
(a) $\sin^{-1} x + k$
(b) $\frac{1}{2}\log\left|\frac{x-1}{x+1}\right| + k$
(c) $\frac{1}{2}\log\left|\frac{x+1}{x-1}\right| + k$
(d) $\sqrt{1 - x^2} + k$
›Reveal solutionSolution

$\int \frac{dx}{x^2-1} = \frac{1}{2}\log\left|\frac{x-1}{x+1}\right| + k$ .

The standard result is $\int \frac{dx}{x^2-a^2} = \frac{1}{2a}\log\left|\frac{x-a}{x+a}\right| + k$ .

…

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CBSE 2021Set ANNUAL1 markMCQ
$\displaystyle\int \dfrac{x\,dx}{(x-1)(x-2)}$
(a) $\log\left|\dfrac{(x-1)^{2}}{x-2}\right|+C$
(b) $\log\left|\dfrac{(x-2)^{2}}{x-1}\right|+C$
(c) $\log\left|\left(\dfrac{x-1}{x-2}\right)^{2}\right|+C$
(d) $\log\left|(x-1)(x-2)\right|+C$
›Reveal solutionSolution

Partial fractions give $\dfrac{x}{(x-1)(x-2)} = \dfrac{-1}{x-1}+\dfrac{2}{x-2}$ , integrating to $\log\left|\dfrac{(x-2)^2}{x-1}\right|+C$ .

Let $\dfrac{x}{(x-1)(x-2)} = \dfrac{A}{x-1}+\dfrac{B}{x-2}$

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

At $x=1$ : $1 = -A \Rightarrow A=-1$

At $x=2$ : $2 = B \Rightarrow B=2$

…

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CBSE 2019Set ANNUAL1 markMCQ
$Q. If 1/(x(x-3)) = A/x + B/(x-3), then the value of B is:$
(a) 1/2
(b) −1/3
(c) 1/3
(d) −1/2
›Reveal solutionSolution

Clear the denominator and compare coefficients (or plug in $x=3$ ) to isolate $B$ .

Write $\dfrac{1}{x(x-3)} = \dfrac{A}{x} + \dfrac{B}{x-3}$ .

Multiplying both sides by $x(x-3)$ :

$1 = A(x-3) + Bx$

…

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CBSE 2019Set ANNUAL1 markMCQ
$Q. The value of \int dx/(x²-a²) is:$
(a) (1/a) tan⁻¹(x/a) + C
(b) (1/2a) log((x−a)/(x+a)) + C
(c) sin⁻¹(x/a) + C
(d) (1/2a) log((x+a)/(x−a)) + C
›Reveal solutionSolution

This is a standard result obtained by partial fractions: $\dfrac{1}{x^2-a^2} = \dfrac{1}{2a}\left(\dfrac{1}{x-a} - \dfrac{1}{x+a}\right)$ .

$\int \frac{dx}{x^2-a^2} = \frac{1}{2a}\int\left(\frac{1}{x-a} - \frac{1}{x+a}\right)dx = \frac{1}{2a}\left[\log|x-a| - \log|x+a|\right] + C$

…

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CBSE 2018Set ANNUAL1 markMCQ
$\frac{1}{(1+\sin x)(2+\sin x)} = \frac{a}{(1+\sin x)} + \frac{b}{(2+\sin x)}$
(a) $0$
(b) $1$
(c) $2$
(d) $3$
›Reveal solutionSolution

Partial fractions give $a=1,\ b=-1$ , hence $a+b=0$ .

Let $t=\sin x$ . Then $\dfrac{1}{(1+t)(2+t)}=\dfrac{a}{1+t}+\dfrac{b}{2+t}$ , so $1=a(2+t)+b(1+t)$ .

…

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