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JEE Mains Maths · Indefinite Integration

Rational Functions and Standard Forms

Splitting a fraction into partial fractions, completing a square to reach a standard integral, and writing a numerator through the denominator and its derivative.

Why this matters

Eleven PYQs, six of them multiple choice, and one from 2026. Four split a fraction into partial fractions, three of them after putting t = x², t = tan x or t = x eˣ; three complete a square or use x ± 1/x to reach a standard form; four write the numerator as a multiple of the denominator plus a multiple of its derivative. Three ideas cover the page.

Concept 1 of 3: Partial fractions

A fraction with a factored denominator is a sum of simpler fractions, one for each factor, and each of those integrates to a log or an inverse tangent. For distinct linear factors, each coefficient is found by covering its factor and putting in the root. Many questions hide the fraction: substitute first, then split.

Definition

  • If the top's degree is not below the bottom's, divide first.
  • Distinct linear factors: Ax−a\frac{A}{x-a} for each, with AA the cover-up value at x=ax=a.
  • A factor that does not split, such as x2+1x^2+1, takes Bx+Cx2+1\frac{Bx+C}{x^2+1}.
  • A function of x2x^2, tan⁡x\tan x or xexxe^x: put tt equal to it first, then split in tt.

Cover-up rule

px+q(x−a)(x−b)=Ax−a+Bx−b,A=pa+qa−b\frac{px+q}{(x-a)(x-b)}=\frac{A}{x-a}+\frac{B}{x-b},\quad A=\frac{pa+q}{a-b}

Worked example

Find ∫3x+5x2−1 dx\int\frac{3x+5}{x^2-1}\,dx.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 2 Apr 2026 Shift 2 · Q69Moderate

Example 1 · Indefinite Integration · Rational Functions and Standard Forms

Let f(x)=∫(16x+24x2+2x−15)dxf(x) = \int\left( \frac{16x+ 24}{x^{2}+ 2x- 15} \right)dx. If f(4)=14log⁡e(3)f(4) = 14\log_{e}(3) and f(7)=log⁡e(2α⋅3β),α,β∈Nf(7) =\log_{e}\left( 2^{\alpha}\cdot3^{\beta} \right),\alpha,\beta\in N, then α+β\alpha+\beta is equal to:

Divide before splitting

Partial fractions need the top's degree below the bottom's. For x2x2−1\frac{x^2}{x^2-1}, first write 1+1x2−11+\frac{1}{x^2-1}; splitting straight away loses the 1.

Concept 2 of 3: Completing the square and x ± 1/x

A quadratic that does not factor is a square plus a constant, and one over that is an inverse tangent. Complete the square first and the standard table does the rest. When the top is x2±1x^2\pm1 and the bottom is a quartic with no odd powers, divide by x2x^2: the top becomes the derivative of x∓1xx\mp\frac1x.

Definition

  • ∫dxx2+a2=1atan⁡−1xa\int\frac{dx}{x^2+a^2}=\frac1a\tan^{-1}\frac xa; ∫dxx2−a2=12aln⁡∣x−ax+a∣\int\frac{dx}{x^2-a^2}=\frac1{2a}\ln\left|\frac{x-a}{x+a}\right|.
  • ∫dxx2+a2=ln⁡∣x+x2+a2∣\int\frac{dx}{\sqrt{x^2+a^2}}=\ln\left|x+\sqrt{x^2+a^2}\right|; ∫dxa2−x2=sin⁡−1xa\int\frac{dx}{\sqrt{a^2-x^2}}=\sin^{-1}\frac xa.
  • ∫x2+a2 dx=x2x2+a2+a22ln⁡∣x+x2+a2∣\int\sqrt{x^2+a^2}\,dx=\frac x2\sqrt{x^2+a^2}+\frac{a^2}2\ln\left|x+\sqrt{x^2+a^2}\right|.
  • For x2±1x4+kx2+1\frac{x^2\pm1}{x^4+kx^2+1}, divide by x2x^2 and put t=x∓1xt=x\mp\frac1x. Higher powers work the same way: t=x3+1x3t=x^3+\frac1{x^3}.

Complete the square

∫dx(x+p)2+a2=1atan⁡−1x+pa+C\int\frac{dx}{(x+p)^2+a^2}=\frac1a\tan^{-1}\frac{x+p}{a}+C

Worked example

Find ∫dxx2+4x+13\int\frac{dx}{x^2+4x+13}.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 27 Jan 2024 · Q164Moderate

Example 2 · Indefinite Integration · Rational Functions and Standard Forms

The integral ∫(x8−x2)dx(x12+3x6+1)tan⁡−1(x3+1x3)\int\frac{\left( x^{8}-x^{2} \right)dx}{\left( x^{12}+ 3x^{6}+ 1 \right)\tan^{- 1}\left( x^{3}+\frac{1}{x^{3}} \right)} is equal to :

Match the sign to the top

For x2+1x^2+1 on top, put t=x−1xt=x-\frac1x; for x2−1x^2-1, put t=x+1xt=x+\frac1x. The other choice leaves no dtdt in the numerator.

Concept 3 of 3: Numerator through the denominator

Write the numerator as a multiple of the denominator plus a multiple of its derivative. The first part integrates to a multiple of xx, and the second to a log. For a quadratic under a root, the same split turns the integral into a root, a log and ∫Q\int\sqrt Q, each from the table.

Definition

  • asin⁡x+bcos⁡xcsin⁡x+dcos⁡x\frac{a\sin x+b\cos x}{c\sin x+d\cos x} or aex+be−xcex+de−x\frac{ae^x+be^{-x}}{ce^x+de^{-x}}: write the top as A g+B g′A\,g+B\,g', where gg is the bottom.
  • Match the coefficients of sin⁡x\sin x and cos⁡x\cos x (or exe^x and e−xe^{-x}) to find AA and BB.
  • px+qQ\frac{px+q}{Q} or px+qQ\frac{px+q}{\sqrt Q}: top =A Q′+B=A\,Q'+B.
  • quadraticQ\frac{\text{quadratic}}{\sqrt Q}: top =A Q+B Q′+C=A\,Q+B\,Q'+C.

Split the numerator

∫A g(x)+B g′(x)g(x) dx=Ax+Bln⁡∣g(x)∣+C\int\frac{A\,g(x)+B\,g'(x)}{g(x)}\,dx=Ax+B\ln|g(x)|+C

Worked example

Find ∫sin⁡xsin⁡x+cos⁡x dx\int\frac{\sin x}{\sin x+\cos x}\,dx.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 24 Jan 2025 · Q150Moderate

Example 3 · Indefinite Integration · Rational Functions and Standard Forms

If ∫2x2+5x+9x2+x+1dx=xx2+x+1+αx2+x+1+\int\frac{2x^{2}+ 5x+ 9}{\sqrt{x^{2}+x+ 1}}dx=x\sqrt{x^{2}+x+ 1}+\alpha\sqrt{x^{2}+x+ 1}+ βlog⁡e∣x+12+x2+x+1∣+C\beta\log_{e}\left| x +\frac{1}{2}+\sqrt{x^{2}+ x + 1} \right|+C, where C is the constant of integration, then α+2β\alpha+ 2\beta is equal to ____\_\_\_\_

Keep the leftover constant

A quadratic top needs three pieces: A Q+B Q′+CA\,Q+B\,Q'+C. With only the first two, the constant CC is lost, and so is its ∫dxQ\int\frac{dx}{\sqrt Q} log term.

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

  • Partial fractions

    Cover-up rule

    px+q(x−a)(x−b)=Ax−a+Bx−b,A=pa+qa−b\frac{px+q}{(x-a)(x-b)}=\frac{A}{x-a}+\frac{B}{x-b},\quad A=\frac{pa+q}{a-b}
  • Completing the square and x ± 1/x

    Complete the square

    ∫dx(x+p)2+a2=1atan⁡−1x+pa+C\int\frac{dx}{(x+p)^2+a^2}=\frac1a\tan^{-1}\frac{x+p}{a}+C
  • Numerator through the denominator

    Split the numerator

    ∫A g(x)+B g′(x)g(x) dx=Ax+Bln⁡∣g(x)∣+C\int\frac{A\,g(x)+B\,g'(x)}{g(x)}\,dx=Ax+B\ln|g(x)|+C

Watch out for (3)

Test yourself on Indefinite Integration

15 past JEE Mains questions from this chapter, timed at 36 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.