Question 273 marksShort AnswerSection C

(a) Find: ∫x+2x−2 dx\int \sqrt{\dfrac{x+2}{x-2}} \, dx.


OR


(b) Find: ∫x2(x2+9)(x2+16) dx\int \dfrac{x^2}{(x^2+9)(x^2+16)} \, dx.

Integrals
Integration of Irrational and Rational Functions

Official answer

Both alternatives are solved with standard integral forms after an algebraic step.


Option (a): x2−4+2ln⁡∣x+x2−4∣+C\sqrt{x^2 - 4} + 2 \ln|x + \sqrt{x^2 - 4}| + C


  • Rationalise: (x+2)/(x−2)=(x+2)/x2−4\sqrt{(x+2)/(x-2)} = (x+2)/\sqrt{x^2-4}.
  • Split: ∫x/x2−4 dx+2∫dx/x2−4=x2−4+2ln⁡∣x+x2−4∣+C\int x/\sqrt{x^2-4} \, dx + 2\int dx/\sqrt{x^2-4} = \sqrt{x^2-4} + 2 \ln|x + \sqrt{x^2-4}| + C.

Option (b): (4/7)tan⁡−1(x/4)−(3/7)tan⁡−1(x/3)+C(4/7) \tan^{-1}(x/4) - (3/7) \tan^{-1}(x/3) + C


  • Put t=x2t = x^2; partial fractions give x2/[(x2+9)(x2+16)]=−(9/7)/(x2+9)+(16/7)/(x2+16)x^2/[(x^2+9)(x^2+16)] = -(9/7)/(x^2+9) + (16/7)/(x^2+16).
  • Integrate each standard form: −(9/7)⋅(1/3)tan⁡−1(x/3)+(16/7)⋅(1/4)tan⁡−1(x/4)-(9/7)\cdot(1/3)\tan^{-1}(x/3) + (16/7)\cdot(1/4)\tan^{-1}(x/4) = (4/7)tan⁡−1(x/4)−(3/7)tan⁡−1(x/3)+C(4/7)\tan^{-1}(x/4) - (3/7)\tan^{-1}(x/3) + C.
Answer keywords:integration by substitutionstandard integral formsconjugate rationalisationpartial fractionstan inverse standard integralirrational function integration

Marking scheme

  1. 1.1 mark: correct simplification step — rationalisation for (a), or partial-fraction decomposition (A=−9/7A = -9/7, B=16/7B = 16/7) for (b).
  2. 2.1 mark: correct application of standard forms ∫x/x2−a2dx\int x/\sqrt{x^2-a^2}dx and ∫dx/x2−a2\int dx/\sqrt{x^2-a^2} (a), or ∫dx/(x2+a2)=(1/a)tan⁡−1(x/a)\int dx/(x^2+a^2) = (1/a)\tan^{-1}(x/a) (b).
  3. 3.1 mark: correct final answer with constant of integration C.

Hint

For (a), rationalise by multiplying inside the root by (x+2)/(x+2)(x+2)/(x+2) to get (x+2)/x2−4(x+2)/\sqrt{x^2-4}. For (b), set t=x2t = x^2 and split x2/[(x2+9)(x2+16)]x^2/[(x^2+9)(x^2+16)] into partial fractions before integrating.

Quick oral answer

Part (a) rationalises to x2−4+2ln⁡∣x+x2−4∣+C\sqrt{x^2-4} + 2\ln|x+\sqrt{x^2-4}| + C; part (b) uses partial fractions (t=x2t = x^2) to reach (4/7)tan⁡−1(x/4)−(3/7)tan⁡−1(x/3)+C(4/7)\tan^{-1}(x/4) - (3/7)\tan^{-1}(x/3) + C.

Common mistakes

  1. 1.Forgetting to rationalise by (x+2)/(x+2)(x+2)/(x+2) inside the square root in part (a), leaving an unintegrable form.
  2. 2.Omitting the constant of integration C at the end of an indefinite integral.
  3. 3.In part (b), errors solving the partial-fraction system A+B=1A + B = 1, 16A+9B=016A + 9B = 0, or dropping the 1/3 and 1/4 factors from the arctangent standard forms.

Analysis and explanation

Both parts of this OR question test recognition of standard integral forms after a simplification step, a core skill in the Integrals chapter.


Concept: Part (a) is a classic irrational-function integral solved by rationalising — multiplying (x+a)/(x−a)\sqrt{(x+a)/(x-a)} inside the root by (x+a)/(x+a)(x+a)/(x+a) converts it to (x+a)/x2−a2(x+a)/\sqrt{x^2-a^2}, which splits into ∫x/x2−a2 dx\int x/\sqrt{x^2-a^2}\,dx and ∫dx/x2−a2\int dx/\sqrt{x^2-a^2}. Part (b) is a rational-function integral solved by partial fractions in t=x2t = x^2, then applying ∫dx/(x2+a2)=(1/a)tan⁡−1(x/a)\int dx/(x^2+a^2) = (1/a)\tan^{-1}(x/a).


Exam trap: In part (a) students often forget the domain restriction (x>2x > 2 or x≤−2x \le -2) or drop the constant of integration. In part (b) they try direct substitution instead of decomposing into partial fractions, or make sign errors solving A+B=1A + B = 1 and 16A+9B=016A + 9B = 0.


Real-world link: Integrals of the form ∫dx/(x²+a²) yielding arctangent appear throughout physics — e.g. the magnetic field of a finite current-carrying wire and RC-circuit phase calculations.

Interesting facts

Integrals of the form ∫dx/x2−a2\int dx/\sqrt{x^2-a^2} are directly linked to the inverse hyperbolic cosine function, cosh⁡−1(x/a)\cosh^{-1}(x/a), an alternative closed form to the logarithmic expression.

Partial fraction decomposition, used in the alternate reading of part (b), was formalised by mathematicians including Johann Bernoulli and Leibniz in the late 17th century for integrating rational functions.

Question details

Q27 of 44 questions

Board
CBSE
Class
Class 12
Subject
Mathematics
Year
2026
Section
Section C
Marks
3
Type
Short Answer
Chapter
Integrals
Topic
Integration of Irrational and Rational Functions
Bloom's level
apply
Syllabus topic
Unit 4: Integrals — Integration of Irrational and Rational Functions (Standard Forms)
Author
CBSE
Reviewed by
Padhantu Academic Team

Spotted a mistake or something unclear?

Tell us and we will fix it.

Frequently asked questions

Why multiply by the conjugate inside the square root in part (a)?

Multiplying (x+2)/(x−2)\sqrt{(x+2)/(x-2)} by (x+2)/(x+2)(x+2)/(x+2) converts the expression into (x+2)/x2−4(x+2)/\sqrt{x^2-4}, which splits neatly into two standard integral forms; without this step the integrand cannot be integrated directly.

How is part (b) integrated?

Substitute t=x2t = x^2 and decompose x2/[(x2+9)(x2+16)]x^2/[(x^2+9)(x^2+16)] into partial fractions −(9/7)/(x2+9)+(16/7)/(x2+16)-(9/7)/(x^2+9) + (16/7)/(x^2+16), then integrate each term using ∫dx/(x2+a2)=(1/a)tan⁡−1(x/a)\int dx/(x^2+a^2) = (1/a)\tan^{-1}(x/a).