What is Integration Using Partial Fractions in Class 12 Maths?

Integration Using Partial Fractions is part of Chapter 7 (Integrals) in NCERT Class 12 Mathematics. This topic is fundamental for CBSE board exams and competitive tests.

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Integration Using Partial Fractions

🎓 Class 12 Mathematics CBSE Theory Ch 7 — Integrals ⏱ ~25 min

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7.4 Integrals of Some Particular Functions

In this section, we mention below some important formulae of integrals and apply them for integrating many other related standard integrals.

Standard Formulae for Rational Functions

#	Integral	Result
1	\(\int \frac{dx}{x^2 - a^2}\)	\(\frac{1}{2a}\log\left\|\frac{x-a}{x+a}\right\| + C\)
2	\(\int \frac{dx}{a^2 - x^2}\)	\(\frac{1}{2a}\log\left\|\frac{a+x}{a-x}\right\| + C\)
3	\(\int \frac{dx}{x^2 + a^2}\)	\(\frac{1}{a}\tan^{-1}\frac{x}{a} + C\)
4	\(\int \frac{dx}{\sqrt{x^2 - a^2}}\)	\(\log\|x + \sqrt{x^2 - a^2}\| + C\)
5	\(\int \frac{dx}{\sqrt{a^2 - x^2}}\)	\(\sin^{-1}\frac{x}{a} + C\)
6	\(\int \frac{dx}{\sqrt{x^2 + a^2}}\)	\(\log\|x + \sqrt{x^2 + a^2}\| + C\)

Proof of Formula 1: \(\int \frac{dx}{x^2 - a^2}\)

Derivation

We have \(\frac{1}{x^2 - a^2} = \frac{1}{(x-a)(x+a)}\). Using partial fractions: \[\frac{1}{(x-a)(x+a)} = \frac{1}{2a}\left[\frac{1}{x-a} - \frac{1}{x+a}\right]\] Therefore: \[\int \frac{dx}{x^2 - a^2} = \frac{1}{2a}\left[\int \frac{dx}{x-a} - \int \frac{dx}{x+a}\right] = \frac{1}{2a}[\log|x-a| - \log|x+a|] + C = \frac{1}{2a}\log\left|\frac{x-a}{x+a}\right| + C\]

Proof of Formula 3: \(\int \frac{dx}{x^2 + a^2}\)

Derivation

Put \(x = a\tan\theta\). Then \(dx = a\sec^2\theta\,d\theta\). \[\int \frac{a\sec^2\theta\,d\theta}{a^2\tan^2\theta + a^2} = \int \frac{a\sec^2\theta}{a^2\sec^2\theta}\,d\theta = \frac{1}{a}\int d\theta = \frac{\theta}{a} + C = \frac{1}{a}\tan^{-1}\frac{x}{a} + C\]

Proof of Formula 5: \(\int \frac{dx}{\sqrt{a^2 - x^2}}\)

Derivation

Put \(x = a\sin\theta\), so \(dx = a\cos\theta\,d\theta\). \[\int \frac{a\cos\theta\,d\theta}{\sqrt{a^2 - a^2\sin^2\theta}} = \int \frac{a\cos\theta}{a\cos\theta}\,d\theta = \int d\theta = \theta + C = \sin^{-1}\frac{x}{a} + C\]

Worked Examples

Example 8

Find the following integrals:

(i) \(\int \frac{dx}{x^2 - 16}\) (ii) \(\int \frac{dx}{\sqrt{2x - x^2}}\)

Solution

(i) \(x^2 - 16 = x^2 - 4^2\). Using formula 1 with \(a = 4\): \[\int \frac{dx}{x^2 - 16} = \frac{1}{2(4)}\log\left|\frac{x-4}{x+4}\right| + C = \frac{1}{8}\log\left|\frac{x-4}{x+4}\right| + C\]

(ii) Complete the square: \(2x - x^2 = -(x^2 - 2x) = -(x^2 - 2x + 1 - 1) = 1 - (x-1)^2\).
Using formula 5 with \(a = 1\) and replacing \(x\) by \(x - 1\): \[\int \frac{dx}{\sqrt{1 - (x-1)^2}} = \sin^{-1}(x-1) + C\]

Example 9

Find \(\int \frac{dx}{\sqrt{x^2 - 2x + 3}}\)

Solution

Complete the square: \(x^2 - 2x + 3 = (x-1)^2 + 2\).
Put \(x - 1 = t\), \(dx = dt\). \[\int \frac{dt}{\sqrt{t^2 + (\sqrt{2})^2}} = \log|t + \sqrt{t^2 + 2}| + C = \log|(x-1) + \sqrt{x^2 - 2x + 3}| + C\]

7.5 Integration by Partial Fractions

Recall that a rational function^? is a ratio of two polynomials of the form \(\frac{P(x)}{Q(x)}\), where \(P(x)\) and \(Q(x)\) are polynomials in \(x\) and \(Q(x) \neq 0\).

Proper vs Improper Rational Function

If the degree of \(P(x)\) is less than the degree of \(Q(x)\), then the rational function is called proper. Otherwise, it is called improper.

If the function is improper, we divide \(P(x)\) by \(Q(x)\) to get: \(\frac{P(x)}{Q(x)} = T(x) + \frac{P_1(x)}{Q(x)}\), where \(T(x)\) is a polynomial and \(\frac{P_1(x)}{Q(x)}\) is a proper rational function.

Table of Partial Fraction Decompositions

S. No.	Form of Rational Function	Partial Fraction Decomposition
1	\(\frac{px + q}{(x-a)(x-b)}\), \(a \neq b\)	\(\frac{A}{x-a} + \frac{B}{x-b}\)
2	\(\frac{px + q}{(x-a)^2}\)	\(\frac{A}{x-a} + \frac{B}{(x-a)^2}\)
3	\(\frac{px^2 + qx + r}{(x-a)(x-b)(x-c)}\)	\(\frac{A}{x-a} + \frac{B}{x-b} + \frac{C}{x-c}\)
4	\(\frac{px^2 + qx + r}{(x-a)^2(x-b)}\)	\(\frac{A}{x-a} + \frac{B}{(x-a)^2} + \frac{C}{x-b}\)
5	\(\frac{px^2 + qx + r}{(x-a)(x^2+bx+c)}\)	\(\frac{A}{x-a} + \frac{Bx+C}{x^2+bx+c}\)

where \(x^2 + bx + c\) cannot be factorised further (i.e., \(b^2 - 4c < 0\)).

Example 10

Find \(\int \frac{x}{(x+1)(x+2)}\,dx\)

Solution

Write \(\frac{x}{(x+1)(x+2)} = \frac{A}{x+1} + \frac{B}{x+2}\).
This gives \(x = A(x+2) + B(x+1)\).
Setting \(x = -1\): \(-1 = A(1) \Rightarrow A = -1\).
Setting \(x = -2\): \(-2 = B(-1) \Rightarrow B = 2\).
Thus: \[\int \frac{x}{(x+1)(x+2)}\,dx = \int \left[\frac{-1}{x+1} + \frac{2}{x+2}\right]\,dx = -\log|x+1| + 2\log|x+2| + C\] \[= \log\left|\frac{(x+2)^2}{x+1}\right| + C\]

Example 11

Find \(\int \frac{x^2 + 1}{x^2 - 5x + 6}\,dx\)

Solution

Since degree of numerator = degree of denominator, this is an improper fraction. Divide first:
\(\frac{x^2 + 1}{x^2 - 5x + 6} = 1 + \frac{5x - 5}{x^2 - 5x + 6} = 1 + \frac{5x - 5}{(x-2)(x-3)}\).

Now decompose: \(\frac{5x - 5}{(x-2)(x-3)} = \frac{A}{x-2} + \frac{B}{x-3}\).
\(5x - 5 = A(x-3) + B(x-2)\).
Setting \(x = 2\): \(5 = A(-1) \Rightarrow A = -5\).
Setting \(x = 3\): \(10 = B(1) \Rightarrow B = 10\).
Therefore: \[\int \frac{x^2 + 1}{x^2 - 5x + 6}\,dx = \int \left[1 - \frac{5}{x-2} + \frac{10}{x-3}\right]\,dx = x - 5\log|x-2| + 10\log|x-3| + C\]

Example 12

Find \(\int \frac{x}{(x-1)^2(x+2)}\,dx\)

Solution

\(\frac{x}{(x-1)^2(x+2)} = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{x+2}\).
So \(x = A(x-1)(x+2) + B(x+2) + C(x-1)^2\).
Setting \(x = 1\): \(1 = 3B \Rightarrow B = \frac{1}{3}\).
Setting \(x = -2\): \(-2 = 9C \Rightarrow C = -\frac{2}{9}\).
Comparing coefficients of \(x^2\): \(0 = A + C \Rightarrow A = \frac{2}{9}\).
Therefore: \[\int \frac{x}{(x-1)^2(x+2)}\,dx = \frac{2}{9}\log|x-1| - \frac{1}{3(x-1)} - \frac{2}{9}\log|x+2| + C = \frac{2}{9}\log\left|\frac{x-1}{x+2}\right| - \frac{1}{3(x-1)} + C\]

Interactive: Partial Fraction Decomposer

Enter values for \(\frac{x + k}{(x - a)(x - b)}\) and see the decomposition

k =

a =

b =

Click Decompose to see the partial fraction decomposition.

Exercise 7.5

Integrate the rational functions in Exercises 1 to 21:

1. \(\int \frac{x}{(x+1)(x+2)}\,dx\)

\(\frac{x}{(x+1)(x+2)} = \frac{-1}{x+1} + \frac{2}{x+2}\). Answer: \(-\log|x+1| + 2\log|x+2| + C\).

2. \(\int \frac{1}{x^2-9}\,dx\)

\(x^2 - 9 = (x-3)(x+3)\). Using formula: \(\frac{1}{6}\log\left|\frac{x-3}{x+3}\right| + C\).

3. \(\int \frac{3x-1}{(x-1)(x-2)(x-3)}\,dx\)

Decompose: \(\frac{A}{x-1} + \frac{B}{x-2} + \frac{C}{x-3}\). Setting \(x = 1\): \(2 = A(1-2)(1-3) = 2A \Rightarrow A = 1\). Setting \(x = 2\): \(5 = B(-1)(1) \Rightarrow B = -5\). Setting \(x = 3\): \(8 = C(2)(1) \Rightarrow C = 4\).
Answer: \(\log|x-1| - 5\log|x-2| + 4\log|x-3| + C\).

4. \(\int \frac{x}{(x-1)(x-2)(x-3)}\,dx\)

Setting \(x = 1\): \(1 = A(-1)(-2) = 2A \Rightarrow A = \frac{1}{2}\). Setting \(x = 2\): \(2 = B(1)(-1) \Rightarrow B = -2\). Setting \(x = 3\): \(3 = C(2)(1) \Rightarrow C = \frac{3}{2}\).
Answer: \(\frac{1}{2}\log|x-1| - 2\log|x-2| + \frac{3}{2}\log|x-3| + C\).

5. \(\int \frac{2x}{x^2+3x+2}\,dx\)

\(x^2+3x+2 = (x+1)(x+2)\). \(\frac{2x}{(x+1)(x+2)} = \frac{-2}{x+1} + \frac{4}{x+2}\).
Answer: \(-2\log|x+1| + 4\log|x+2| + C\).

6. \(\int \frac{1-x^2}{x(1-2x)}\,dx\)

This is improper. Dividing: \(\frac{1-x^2}{x(1-2x)} = \frac{1-x^2}{x - 2x^2}\). Long division gives: \(\frac{1}{2} + \frac{1-\frac{x}{2}}{x(1-2x)}\). Using partial fractions and integrating: \(\frac{x}{2} + \frac{1}{2}\log|x| + \frac{1}{4}\log|1-2x| + C\).

7. \(\int \frac{x}{(x^2+1)(x-1)}\,dx\)

\(\frac{x}{(x^2+1)(x-1)} = \frac{A}{x-1} + \frac{Bx+C}{x^2+1}\). Setting \(x = 1\): \(1 = 2A \Rightarrow A = \frac{1}{2}\). Comparing: \(B = -\frac{1}{2}\), \(C = \frac{1}{2}\).
Answer: \(\frac{1}{2}\log|x-1| - \frac{1}{4}\log(x^2+1) + \frac{1}{2}\tan^{-1}x + C\).

8. \(\int \frac{x}{(x-1)^2(x+2)}\,dx\)

Already solved in Example 12: \(\frac{2}{9}\log\left|\frac{x-1}{x+2}\right| - \frac{1}{3(x-1)} + C\).

Choose the correct answer in Exercises 22 and 23:

22. \(\int \frac{x\,dx}{(x-1)(x-2)}\) equals:
(A) \(\log\left|\frac{(x-1)^2}{x-2}\right| + C\) (B) \(\log\left|\frac{(x-2)^2}{x-1}\right| + C\)
(C) \(\log\left|\left(\frac{x-1}{x-2}\right)^2\right| + C\) (D) \(\log|(x-1)(x-2)| + C\)

\(\frac{x}{(x-1)(x-2)} = \frac{A}{x-1} + \frac{B}{x-2}\). \(x = A(x-2) + B(x-1)\). At \(x = 1\): \(A = -1\). At \(x = 2\): \(B = 2\).
\(-\log|x-1| + 2\log|x-2| + C = \log\left|\frac{(x-2)^2}{x-1}\right| + C\). Answer: (B)

23. \(\int \frac{dx}{x(x^2+1)}\) equals:
(A) \(\log|x| - \frac{1}{2}\log(x^2+1) + C\) (B) \(\log|x| + \frac{1}{2}\log(x^2+1) + C\)
(C) \(-\log|x| + \frac{1}{2}\log(x^2+1) + C\) (D) \(\frac{1}{2}\log|x| + \log(x^2+1) + C\)

\(\frac{1}{x(x^2+1)} = \frac{1}{x} - \frac{x}{x^2+1}\). Integrating: \(\log|x| - \frac{1}{2}\log(x^2+1) + C\). Answer: (A)

Activity: Complete the Square Practice

L3 Apply

Predict: Can you rewrite each quadratic expression in completed-square form?

Write each in the form \((x \pm h)^2 \pm k\): (a) \(x^2 + 6x + 5\) (b) \(x^2 - 4x + 8\) (c) \(3 - 2x - x^2\)
Identify which standard formula (1 through 6) applies to \(\int \frac{dx}{\text{expression}}\).
Write the integral without computing it.

(a) \((x+3)^2 - 4\) → Formula 1 with \(a = 2\), substitution \(t = x+3\).

(b) \((x-2)^2 + 4\) → Formula 3 with \(a = 2\), substitution \(t = x-2\).

Competency-Based Questions

Scenario: An electrical engineer analyses a circuit where the current \(I(t)\) through a component satisfies \(\frac{dI}{dt} = \frac{5}{(t+1)(t+3)}\) amperes/second. The initial current is \(I(0) = 0\).

Q1. The partial fraction decomposition of \(\frac{5}{(t+1)(t+3)}\) is:

L2 Understand

(a) \(\frac{5/2}{t+1} - \frac{5/2}{t+3}\)
(b) \(\frac{5}{t+1} + \frac{5}{t+3}\)
(c) \(\frac{5}{t+1} - \frac{5}{t+3}\)
(d) \(\frac{1}{t+1} - \frac{1}{t+3}\)

Answer: (a). \(\frac{5}{(t+1)(t+3)} = \frac{A}{t+1} + \frac{B}{t+3}\). \(5 = A(t+3) + B(t+1)\). At \(t = -1\): \(A = \frac{5}{2}\). At \(t = -3\): \(B = -\frac{5}{2}\).

Q2. Find \(I(t)\) and compute the current at \(t = 2\) seconds.

L3 Apply

Answer: \(I(t) = \frac{5}{2}[\log(t+1) - \log(t+3)] + C = \frac{5}{2}\log\frac{t+1}{t+3} + C\). Using \(I(0) = 0\): \(0 = \frac{5}{2}\log\frac{1}{3} + C \Rightarrow C = \frac{5}{2}\log 3\). So \(I(t) = \frac{5}{2}\log\frac{3(t+1)}{t+3}\). At \(t = 2\): \(I(2) = \frac{5}{2}\log\frac{9}{5} = \frac{5}{2}\log 1.8 \approx 1.47\) A.

Q3. As \(t \to \infty\), what value does \(I(t)\) approach? Interpret this physically.

L4 Analyse

Answer: As \(t \to \infty\): \(\frac{3(t+1)}{t+3} \to 3\). So \(I \to \frac{5}{2}\log 3 \approx 2.75\) A. The current approaches a steady-state value of \(\frac{5}{2}\ln 3\) amperes because the rate of change \(\frac{dI}{dt} \to 0\) as \(t \to \infty\).

Q4. If the denominator were \((t+1)^2(t+3)\) instead, how would the partial fraction form change? Write the new form without solving for constants.

L6 Create

Answer: The new decomposition would be: \(\frac{5}{(t+1)^2(t+3)} = \frac{A}{t+1} + \frac{B}{(t+1)^2} + \frac{C}{t+3}\). This follows Type 4 from the partial fractions table — a repeated linear factor \((t+1)^2\) requires two terms.

Assertion–Reason Questions

Assertion (A): \(\int \frac{dx}{x^2 - 4} = \frac{1}{4}\log\left|\frac{x-2}{x+2}\right| + C\).
Reason (R): \(\int \frac{dx}{x^2 - a^2} = \frac{1}{2a}\log\left|\frac{x-a}{x+a}\right| + C\).

(a) Both true, R explains A.

(b) Both true, R doesn't explain A.

(c) A true, R false.

(d) A false, R true.

Answer: (a) — Both true. Here \(a = 2\), so \(\frac{1}{2(2)} = \frac{1}{4}\). R is the general formula, and applying it with \(a = 2\) gives A directly.

Assertion (A): To integrate \(\frac{x^3 + 1}{x^2 - x}\), we must first perform polynomial long division.
Reason (R): Partial fraction decomposition only applies to proper rational functions.

(a) Both true, R explains A.

(b) Both true, R doesn't explain A.

(c) A true, R false.

(d) A false, R true.

Answer: (a) — Both true. Since degree(numerator) = 3 > degree(denominator) = 2, it is improper. We must divide first to get a polynomial plus a proper fraction. R explains why this step is necessary before decomposition.

Assertion (A): \(\frac{1}{(x-1)(x^2+1)} = \frac{A}{x-1} + \frac{B}{x^2+1}\).
Reason (R): When the denominator has an irreducible quadratic factor \(x^2+bx+c\), the corresponding numerator in the partial fraction must be linear (\(Bx + C\)).

(a) Both true, R explains A.

(b) Both true, R doesn't explain A.

(c) A true, R false.

(d) A false, R true.

Answer: (d) — A is false. The correct form is \(\frac{A}{x-1} + \frac{Bx+C}{x^2+1}\), not \(\frac{B}{x^2+1}\). R is true and precisely identifies the error in A: the numerator over an irreducible quadratic must be linear (\(Bx + C\)), not just a constant.

Frequently Asked Questions

What is integration by partial fractions?

Partial fractions decomposes a rational function P(x)/Q(x) into simpler fractions with linear or quadratic denominators that can be integrated individually using standard formulas.

What are the different cases of partial fraction decomposition?

Non-repeated linear factors give A/(x-a) terms. Repeated linear factors give A/(x-a) + B/(x-a)^2 terms. Non-repeated quadratic factors give (Ax+B)/(x^2+bx+c) terms.

How to find partial fraction coefficients?

Multiply both sides by the denominator and compare coefficients, or substitute specific x-values (often roots of the denominator) to solve for constants directly.

When do you need polynomial long division first?

Perform long division when the numerator degree is greater than or equal to the denominator degree. Apply partial fractions only to the resulting proper fraction.

What standard integrals are used with partial fractions?

After decomposition: integral of 1/(x-a) = log|x-a| + C, integral of 1/(x-a)^2 = -1/(x-a) + C, and quadratic forms leading to tan-inverse or log.

Frequently Asked Questions — Integrals

What is Integration Using Partial Fractions in NCERT Class 12 Mathematics?

Integration Using Partial Fractions is a key concept covered in NCERT Class 12 Mathematics, Chapter 7: Integrals. This lesson builds the student's foundation in the chapter by explaining the core ideas with worked examples, definitions, and step-by-step methods aligned to the CBSE curriculum.

How do I solve problems on Integration Using Partial Fractions step by step?

To solve problems on Integration Using Partial Fractions, follow the NCERT method: identify the given quantities, choose the relevant formula or theorem, substitute values carefully, and simplify. Class 12 exercises gradually increase in difficulty — start with solved NCERT examples before attempting exercise questions, and always verify your answer by substitution or diagram.

What are the most important formulas for Chapter 7: Integrals?

The essential formulas of Chapter 7 (Integrals) are listed in the chapter summary and highlighted throughout the lesson in formula boxes. Memorise them and practise at least 2–3 problems per formula. CBSE board exams frequently test direct application as well as combined use of multiple formulas from this chapter.

Is Integration Using Partial Fractions important for the Class 12 board exam?

Integration Using Partial Fractions is part of the NCERT Class 12 Mathematics syllabus and appears in CBSE board exams. Questions typically include short-answer, long-answer, and competency-based items. Review the NCERT examples, exercise questions, and previous-year board problems on this topic to prepare confidently.

What mistakes should students avoid in Integration Using Partial Fractions?

Common mistakes in Integration Using Partial Fractions include skipping steps, misapplying formulas, sign errors, and losing track of units. Write each step clearly, double-check algebraic manipulations, and re-read the question after solving to verify that your answer matches what was asked.

Where can I find more NCERT practice questions on Integration Using Partial Fractions?

End-of-chapter NCERT exercises for Integration Using Partial Fractions cover all difficulty levels tested in CBSE exams. After completing them, try the examples again without looking at the solutions, attempt the NCERT Exemplar questions for Chapter 7, and solve at least one previous-year board paper to consolidate your understanding.

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