MHT-CET Maths · Indefinite Integration

Rational Functions and Partial Fractions

Break a rational integrand into simple standard pieces — arctan/log quadratic forms, completed squares, and partial fractions.

All Indefinite Integration notes MHT-CET Maths strategy

Why this matters

16 PYQs. Four recognitions cover them: the standard quadratic forms (arctan, log, arcsin); completing the square to reach an arctan/arcsin/log; the numerator-split (px+q over a quadratic or its root); and partial-fraction decomposition for products of linear/quadratic factors. MHT-CET also hides quadratic-in-x² shapes (like 1/(x⁴+9x²+16)) that reduce to an arctan after a clever x + k/x substitution.

Concept 1 of 4

Standard Quadratic Denominator Forms

Intuition

A short table of quadratic-denominator integrals produces every arctan and quadratic-log answer. Memorise the shapes; most rational integrals are engineered to land on one of them.

Definition

The core forms the bank draws on:

$\displaystyle\int \dfrac{dx}{x^2 + a^2} = \dfrac{1}{a}\tan^{-1}\dfrac{x}{a} + C$
$\displaystyle\int \dfrac{dx}{x^2 - a^2} = \dfrac{1}{2a}\log\left|\dfrac{x-a}{x+a}\right| + C$
$\displaystyle\int \dfrac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}\dfrac{x}{a} + C$
$\displaystyle\int \dfrac{dx}{\sqrt{x^2 - a^2}} = \log\left|x + \sqrt{x^2 - a^2}\right| + C$
$\displaystyle\int \dfrac{dx}{\sqrt{x^2 + a^2}} = \log\left|x + \sqrt{x^2 + a^2}\right| + C$

Quadratic-in- $x^2$ denominators often reduce to the first form via a $t = x \pm \dfrac{k}{x}$ substitution.

The arctan form

\int \dfrac{dx}{x^2 + a^2} = \dfrac{1}{a}\tan^{-1}\dfrac{x}{a} + C

Worked example

Evaluate

\displaystyle\int \dfrac{x^2 + 1}{x^4 - x^2 + 1}\,dx

Divide top and bottom by $x^2$ : $\dfrac{1 + 1/x^2}{x^2 - 1 + 1/x^2}$ .
Note $\dfrac{d}{dx}\!\left(x - \dfrac{1}{x}\right) = 1 + \dfrac{1}{x^2}$ — exactly the numerator. Let $t = x - \dfrac{1}{x}$ .
Then $t^2 = x^2 - 2 + \dfrac{1}{x^2}$ , so $x^2 + \dfrac{1}{x^2} = t^2 + 2$ and the denominator $= t^2 + 2 - 1 = t^2 + 1$ .
Integral $= \int \dfrac{dt}{t^2 + 1} = \tan^{-1}t = \tan^{-1}\!\left(x - \dfrac{1}{x}\right) + C$ .

Answer:

\tan^{-1}\!\left(x - \dfrac{1}{x}\right) + C

Practice this conceptself-check · 4 quick reps

Try it yourself

Evaluate

\displaystyle\int \dfrac{dx}{x^2 + 9}

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
$\int \dfrac{dx}{x^2+4}$
2.
$\int \dfrac{dx}{\sqrt{9-x^2}}$
3.
$\int \dfrac{dx}{x^2-1}$
4.
$\int \dfrac{dx}{1+x^2}$

From the bank · past-year question

Example 1Indefinite IntegrationHARD

\int \frac{x^2-4}{x^4+9x^2+16}\,dx = \tan^{-1}(f(x)) + c

(where

c

is a constant of integration), then value of

f(2)

[Q113 · 16th May Shift 2 · 2023]

x⁴ + bx² + c → try t = x ± k/x

A denominator quadratic in

x^2

with a matching numerator

(1 \mp k/x^2)

is the signal for

t = x \pm k/x

. It collapses the quartic to a simple

t^2 + 1

arctan — a recurring MHT-CET shape.

Concept 2 of 4

Completing the Square

Intuition

A general quadratic under a root or in a denominator is turned into 'constant ± (linear)²' by completing the square. That matches it to an arcsin, arctan, or log standard form.

Definition

Rewrite $ax^2 + bx + c$ as $a\!\left(x + \dfrac{b}{2a}\right)^2 + \left(c - \dfrac{b^2}{4a}\right)$ . Then $\displaystyle\int \dfrac{dx}{\sqrt{k^2 - (x+p)^2}} = \sin^{-1}\dfrac{x+p}{k}$ , and $\displaystyle\int \dfrac{dx}{(x+p)^2 + k^2} = \dfrac{1}{k}\tan^{-1}\dfrac{x+p}{k}$ .

Completing the square

ax^2 + bx + c = a\left(x + \dfrac{b}{2a}\right)^2 + \left(c - \dfrac{b^2}{4a}\right)

Worked example

Evaluate

\displaystyle\int \dfrac{dx}{\sqrt{5 - 4x - x^2}}

Complete the square inside the root: $5 - 4x - x^2 = -(x^2 + 4x - 5) = -\big((x+2)^2 - 9\big) = 9 - (x+2)^2$ .
Match the arcsin form with $k = 3$ : $\int \dfrac{dx}{\sqrt{9 - (x+2)^2}}$ .
Integrate: $\sin^{-1}\dfrac{x+2}{3} + C$ .

Answer:

\sin^{-1}\dfrac{x+2}{3} + C

Practice this conceptself-check · 4 quick reps

Try it yourself

Evaluate

\displaystyle\int \dfrac{dx}{7 + 6x - x^2}

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Complete the square: $x^2 + 6x$ .
2.
Complete the square: $7 - 6x - x^2$ .
3.
$\int \dfrac{dx}{x^2+2x+5}$
4.
$\int \dfrac{dx}{\sqrt{4-(x-1)^2}}$

From the bank · past-year question

Example 2Indefinite IntegrationHARD

\int\frac{dx}{\sqrt{7-6x-x^2}}=

[Q145 · 10th May Shift 1 · 2024]

Factor the sign on x² before completing the square

With

-x^2

terms, pull out the

-1

first:

7 - 6x - x^2 = -(x^2 + 6x - 7)

. Forgetting the sign flip turns an arcsin into a (wrong) log or vice versa.

Drill 1 more on completing the square

Concept 3 of 4

Linear Numerator over a Quadratic (Numerator Split)

Intuition

When the numerator is linear (px+q) and the denominator is a quadratic or the root of one, split the numerator into 'a constant times the denominator's derivative, plus a leftover constant'. The first piece integrates to a √ or a log; the leftover becomes a complete-the-square standard form.

Definition

For $\displaystyle\int \dfrac{px+q}{\sqrt{ax^2+bx+c}}\,dx$ (or with the quadratic itself in the denominator), write $px + q = A\dfrac{d}{dx}(ax^2+bx+c) + B$ . Match coefficients to find $A, B$ . Then $\displaystyle\int \dfrac{A\,(ax^2+bx+c)'}{\sqrt{ax^2+bx+c}}\,dx = 2A\sqrt{ax^2+bx+c}$ , and the leftover $\displaystyle B\!\int \dfrac{dx}{\sqrt{ax^2+bx+c}}$ is finished by completing the square.

Numerator as derivative-of-denominator plus constant

px + q = A\dfrac{d}{dx}(ax^2+bx+c) + B

Acoefficient that reproduces the $x$ -term via $(ax^2+bx+c)'$
Bleftover constant — its integral completes the square

Worked example

Evaluate

\displaystyle\int \dfrac{3x+2}{\sqrt{x^2+4x+5}}\,dx

Denominator $D = x^2+4x+5$ , so $D' = 2x+4$ . Write $3x+2 = A(2x+4) + B$ .
Match: $x$ -coefficient $3 = 2A\Rightarrow A = \tfrac32$ ; constant $2 = 4A + B = 6 + B\Rightarrow B = -4$ .
First piece: $\displaystyle \tfrac32\!\int \dfrac{D'}{\sqrt{D}}\,dx = \tfrac32\cdot 2\sqrt{D} = 3\sqrt{x^2+4x+5}$ .
Leftover: $\displaystyle -4\!\int \dfrac{dx}{\sqrt{(x+2)^2+1}} = -4\log\left|(x+2)+\sqrt{x^2+4x+5}\right|$ (completing the square).

Answer:

3\sqrt{x^2+4x+5} - 4\log\left|(x+2)+\sqrt{x^2+4x+5}\right| + C

Practice this conceptself-check · 4 quick reps

Try it yourself

Given

\displaystyle\int \dfrac{x-7}{\sqrt{x^2-16x+63}}\,dx = A\sqrt{x^2-16x+63} + \log\left|(x-8)+\sqrt{x^2-16x+63}\right| + c

, find

A

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
For $\int \dfrac{2x+3}{x^2+3x+1}\,dx$ , what is the numerator as $A D' + B$ ?
2.
$\int \dfrac{2x+3}{x^2+3x+1}\,dx$
3.
For $\int \dfrac{x}{\sqrt{x^2+4}}\,dx$ , split $x = A(2x) + B$ .
4.
$\int \dfrac{x}{\sqrt{x^2+4}}\,dx$

From the bank · past-year question

Example 3Indefinite IntegrationHARD

\int\frac{2x+5}{\sqrt{7-6x-x^2}}\,dx = A\sqrt{7-6x-x^2}+B\sin^{-1}\left(\frac{x+3}{4}\right)+c

(where c is a constant of integration) then the value of A+B is

[Q136 · 11th May Shift 2 · 2023]

Split the numerator BEFORE completing the square

The derivative-piece must come out first (it gives the

\sqrt{}

or log term). Only the leftover constant goes through completing the square. Completing the square first leaves the

x

in the numerator with nowhere to go.

Drill 1 more on linear numerator over a quadratic (numerator split)

Concept 4 of 4

Partial-Fraction Decomposition

Intuition

Split a proper rational function into a sum of fractions with linear or simple-quadratic denominators. Each piece is then a basic log or arctan. Divide first if the fraction is improper.

Definition

For a proper rational function, decompose by the denominator's factors:

distinct linear $(x-a)(x-b)$ : $\dfrac{A}{x-a} + \dfrac{B}{x-b}$
repeated linear $(x-a)^2$ : $\dfrac{A}{x-a} + \dfrac{B}{(x-a)^2}$
irreducible quadratic $(x^2 + c)$ : $\dfrac{Ax + B}{x^2 + c}$

Find the constants by the cover-up method or by equating coefficients; integrate each piece.

Distinct-linear decomposition

\dfrac{px + q}{(x-a)(x-b)} = \dfrac{A}{x-a} + \dfrac{B}{x-b}

Worked example

Evaluate

\displaystyle\int \dfrac{2x+1}{x(x+1)}\,dx

Decompose: $\dfrac{2x+1}{x(x+1)} = \dfrac{A}{x} + \dfrac{B}{x+1}$ .
Cover-up at $x = 0$ : $A = \dfrac{0+1}{0+1} = 1$ . At $x = -1$ : $B = \dfrac{2(-1)+1}{-1} = 1$ .
Integrate: $\int\!\left(\dfrac{1}{x} + \dfrac{1}{x+1}\right)dx = \log|x| + \log|x+1| + C$ .

Answer:

\log|x(x+1)| + C

Practice this conceptself-check · 4 quick reps

Try it yourself

Evaluate

\displaystyle\int \dfrac{3x - 2}{(x+1)(x-2)^2}\,dx

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Decompose $\dfrac{1}{(x-1)(x+1)}$ .
2.
$\int \dfrac{dx}{(x-1)(x-2)}$
3.
Improper $\dfrac{x^2}{x^2-1}$ : first step?
4.
Form for $\dfrac{1}{(x-1)^2}$ — already simple. $\int$ ?

From the bank · past-year question

Example 4Indefinite IntegrationMODERATE

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

[Shift || · 2025]

Improper fraction? Divide before decomposing

Partial fractions require the numerator degree to be LESS than the denominator's. If not, polynomial-divide first, then decompose the proper remainder.

Repeated factor needs every power

For

(x-a)^2

you must include BOTH

\dfrac{A}{x-a}

and

\dfrac{B}{(x-a)^2}

. Dropping the first-power term gives an unsolvable system.

Drill 10 more on partial-fraction decomposition

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 (4)

Standard Quadratic Denominator Forms
The arctan form
$\int \dfrac{dx}{x^2 + a^2} = \dfrac{1}{a}\tan^{-1}\dfrac{x}{a} + C$
Completing the Square
Completing the square
$ax^2 + bx + c = a\left(x + \dfrac{b}{2a}\right)^2 + \left(c - \dfrac{b^2}{4a}\right)$
Linear Numerator over a Quadratic (Numerator Split)
Numerator as derivative-of-denominator plus constant
$px + q = A\dfrac{d}{dx}(ax^2+bx+c) + B$
Partial-Fraction Decomposition
Distinct-linear decomposition
$\dfrac{px + q}{(x-a)(x-b)} = \dfrac{A}{x-a} + \dfrac{B}{x-b}$

Watch out for (5)

x⁴ + bx² + c → try t = x ± k/x
→ Standard Quadratic Denominator Forms
Factor the sign on x² before completing the square
→ Completing the Square
Split the numerator BEFORE completing the square
→ Linear Numerator over a Quadratic (Numerator Split)
Improper fraction? Divide before decomposing
→ Partial-Fraction Decomposition
Repeated factor needs every power
→ Partial-Fraction Decomposition

Mastery check — 5 interleaved questions

Try each one before clicking. Questions are interleaved across the concepts above, not grouped — interleaving sharpens transfer.

Example 1Indefinite IntegrationMODERATE

The value of

\int \frac{dx}{7+6x-x^{2}}

is equal to

[Q103 · 15th May Shift 1 · 2023]

Example 2Indefinite IntegrationHARD

\int\frac{x-7}{\sqrt{x^2-16x+63}}\,dx=A\sqrt{x^2-16x+63}+\log|(x-8)+\sqrt{x^2-16x+63}|+c

. Value of A is

[Q112 · 10th May Shift 1 · 2024]

Example 3Indefinite IntegrationHARD

\int\frac{2x^2-1}{(x^2+4)(x^2-3)}\,dx=

[Q102 · 3rd May 2nd Shift · 2023]

Example 4Indefinite IntegrationMODERATE

\int \frac{x^3-7x+6}{x^2+3x}\,dx =

[Q146 · 16th May Shift 2 · 2023]

Example 5Indefinite IntegrationMODERATE

\int \frac{x^3 - 7x + 6}{x^2 + 3x}\,dx =

[Q146 · Shift 1 · 2023]

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