Integrate the following function: $\frac{2}{(1 - x)(1 + x^2)}$

Integrate the following function: (2)/((1 - x)(1 + – Class 12 Mathematics

Exercise 7.5 · Q13

\frac{2}{(1 - x)(1 + x^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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Decompose into a linear piece plus a piece over $1+x^2$ ; integrating gives $-\log|1-x|+\dfrac{1}{2}\log(1+x^2)+\tan^{-1}x+C$ .

Setting up the decomposition

The denominator is a linear factor $(1-x)$ times the irreducible quadratic $(1+x^2)$ (it has no real roots). An irreducible quadratic gets a linear numerator, so

\frac{2}{(1-x)(1+x^2)}=\frac{A}{1-x}+\frac{Bx+C}{1+x^2}.

Solving for the constants

Multiply through by $(1-x)(1+x^2)$ :

2=A(1+x^2)+(Bx+C)(1-x).

Expand the right side: $A+Ax^2+Bx-Bx^2+C-Cx=(A-B)x^2+(B-C)x+(A+C)$ .

Since the left side is $2=0\cdot x^2+0\cdot x+2$ , match coefficients:

A-B=0,\qquad B-C=0,\qquad A+C=2.

The first two give $A=B=C$ , and then $A+C=2$ gives $2A=2$ , so $A=B=C=1$ . (Check: putting $x=1$ in the cleared equation gives $2=2A$ , confirming $A=1$ .)

The decomposed form …

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

AHSEC Higher Secondary (HS) Final Examination 2024Set ANNUAL4 marks
$\displaystyle\int \dfrac{x^2+1}{x^2-5x+6}\, dx$
›Reveal solutionSolution
Reduce the improper rational integrand by division then partial fractions; the OR integral evaluates to $\pi$ using the King's-rule symmetry $x\to\pi-x$ .

Main question: $x^2-5x+6=(x-2)(x-3)$ . Since the numerator and denominator have the same degree, divide first:

$\frac{x^2+1}{x^2-5x+6}=1+\frac{(x^2+1)-(x^2-5x+6)}{x^2-5x+6}=1+\frac{5x-5}{(x-2)(x-3)}.$

Partial fractions: $\dfrac{5x-5}{(x-2)(x-3)}=\dfrac{A}{x-2}+\dfrac{B}{x-3}$ , so $5x-5=A(x-3)+B(x-2)$ .

$x=2$ : $5=-A\Rightarrow A=-5$

$x=3$ : $10=B\Rightarrow B=10$

So the integrand is $1-\dfrac{5}{x-2}+\dfrac{10}{x-3}$ , and

$\int\frac{x^2+1}{x^2-5x+6}dx = x-5\ln|x-2|+10\ln|x-3|+C.$

OR: $I=\displaystyle\int_0^\pi\frac{x\,dx}{1+\sin x}$ . Using $\int_0^a f(x)dx=\int_0^a f(a-x)dx$ with $\sin(\pi-x)=\sin x$ :

$I=\int_0^\pi\frac{(\pi-x)\,dx}{1+\sin x}=\pi\int_0^\pi\frac{dx}{1+\sin x}-I\ \Rightarrow\ 2I=\pi\int_0^\pi\frac{dx}{1+\sin x}.$ …
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AHSEC Higher Secondary (HS) Final Examination 2023Set ANNUAL4 marks
$Q. Integrate (any one):$
(i) $\displaystyle\int \dfrac{2x}{x^2+3x+2}\,dx$
(ii) $\displaystyle\int_0^{2/3} \dfrac{dx}{4+9x^2}$ .
›Reveal solutionSolution

Write the numerator as (derivative of denominator) minus a constant, split the integral, and use partial fractions for the remaining rational part.

(i) $\displaystyle\int \frac{2x}{x^2+3x+2}dx$ :

Note $\dfrac{d}{dx}(x^2+3x+2) = 2x+3$ . Write $2x = (2x+3) - 3$ :

$\int \frac{2x}{x^2+3x+2}dx = \int\frac{2x+3}{x^2+3x+2}dx - 3\int\frac{dx}{x^2+3x+2}.$

The first integral is $\log|x^2+3x+2|$ directly (numerator is the derivative of the denominator).

For the second, factor $x^2+3x+2=(x+1)(x+2)$ , and use partial fractions:

$\frac{1}{(x+1)(x+2)} = \frac{1}{x+1} - \frac{1}{x+2} \implies \int\frac{dx}{x^2+3x+2} = \log|x+1| - \log|x+2|.$

Combining:

$\int\frac{2x}{x^2+3x+2}dx = \log|x^2+3x+2| - 3\left[\log|x+1|-\log|x+2|\right] + C.$

Since $x^2+3x+2 = (x+1)(x+2)$ , $\log|x^2+3x+2| = \log|x+1| + \log|x+2|$ , so

$= \log|x+1|+\log|x+2| - 3\log|x+1| + 3\log|x+2| + C = -2\log|x+1| + 4\log|x+2| + C.$

…

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Q.Integrate the following function: $\frac{2}{(1 - x)(1 + x^2)}$

Concept understanding — Partial Fraction Decomposition

Partial Fraction Decomposition

When it applies

The standard forms

Worked example

Setting up the decomposition

Solving for the constants

The decomposed form …

Method: Partial Fractions with an Irreducible Quadratic Factor

Steps

Common Mistakes

Q.Integrate the following function: 2(1−x)(1+x2)\frac{2}{(1 - x)(1 + x^2)}(1−x)(1+x2)2​

Setting up the decomposition

Solving for the constants

The decomposed form …

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Q.Integrate the following function: $\frac{2}{(1 - x)(1 + x^2)}$

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