MHT-CET Maths · Indefinite Integration

Integration by Substitution — the Workhorse Method

Spot an inner function whose derivative also appears in the integrand, substitute to rename it, and the integral collapses to a standard form.

All Indefinite Integration notes MHT-CET Maths strategy

Why this matters

44 PYQs — by far the largest bucket in the chapter, and the method every other technique falls back on. The single most-tested pattern is f'(x)/f(x) → log|f(x)|. Beyond that: powers of a function times its derivative, root substitutions, and exponential substitutions. Difficulty is steep here (about 60% HARD), but every one of them reduces to 'find u, find du, rewrite, integrate'.

Concept 1 of 6

The Substitution Rule

Intuition

If the integrand contains some inner function

g(x)

AND a factor that looks like its derivative

g'(x)

, substitute

u = g(x)

. The

g'(x)\,dx

becomes

du

, and the whole integral turns into something standard in

u

Definition

If $u = g(x)$ , then $du = g'(x)\,dx$ , and $\int f(g(x))\,g'(x)\,dx = \int f(u)\,du$ . The art is choosing $g(x)$ so that its derivative is already present (up to a constant) in the integrand.

Substitution rule

\int f(g(x))\,g'(x)\,dx = \int f(u)\,du,\quad u = g(x)

u = g(x)the inner function you rename
du = g'(x)\,dxits differential, which must appear in the integrand

Worked example

Evaluate

\displaystyle\int \dfrac{\sin(1/x)}{x^2}\,dx

Let $u = \dfrac{1}{x}$ . Then $du = -\dfrac{1}{x^2}\,dx$ , so $\dfrac{dx}{x^2} = -\,du$ .
Rewrite: $\int \sin u \,(-du) = -\int \sin u\,du$ .
Standard form: $-(-\cos u) + C = \cos u + C$ .
Back-substitute $u = 1/x$ .

Answer:

\cos\dfrac{1}{x} + C

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 1Indefinite IntegrationMODERATE

\int\frac{\tan(1/x)}{x^2}\,dx =

[Q125 · 13th May Shift 1 · 2024]

Adjust for the missing constant

du = g'(x)\,dx

appears up to a numeric factor (e.g. you have

x\,dx

but

du = 2x\,dx

), pull the constant out:

x\,dx = \tfrac12\,du

. Forgetting the

\tfrac12

is the most common slip.

Drill 15 more on the substitution rule

Concept 2 of 6

The f'(x)/f(x) → log Pattern

Intuition

Whenever the numerator is exactly the derivative of the denominator, the integral is just

\log

of the denominator. This is the single highest-yield pattern in the chapter — train your eye to spot 'top = derivative of bottom'.

Definition

If the integrand is a fraction whose numerator is the derivative of its denominator, then $\int \dfrac{f'(x)}{f(x)}\,dx = \log|f(x)| + C$ . Often you must engineer the numerator: split it into 'a constant times $f'(x)$ ' plus a leftover, then integrate each piece.

Logarithmic integral

\int \dfrac{f'(x)}{f(x)}\,dx = \log|f(x)| + C

f(x)the denominator
f'(x)its derivative — must equal the numerator (up to a constant)

Worked example

Evaluate

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

The denominator is $f = x^2 + 3x + 5$ ; its derivative is $f' = 2x + 3$ — exactly the numerator.
So the integrand has the form $f'/f$ , which integrates to $\log|f|$ .
$\int \dfrac{2x+3}{x^2+3x+5}\,dx = \log|x^2+3x+5| + C$ .

Answer:

\log|x^2 + 3x + 5| + C

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 2Indefinite IntegrationHARD

\int\frac{4e^x-25}{2e^x-5}dx = Ax+B\log|2e^x-5|+c

(where c is a constant of integration) then

[Q123 · 4th May Shift 2 · 2023]

Engineer the numerator into f'(x) + leftover

Rarely is the top exactly

f'(x)

. Write it as 'constant

\times f'(x)

+ remainder', send the first piece to a clean log, and handle the remainder separately. The whole-number coefficient comes from matching.

Concept 3 of 6

Power of a Function times its Derivative

Intuition

When you see some function raised to a power, multiplied by (a constant times) its derivative, substitution turns it into the plain power rule. The denominator-squared shapes

\int f'/f^2

are the same idea with a negative power.

Definition

If $u = f(x)$ then $\int [f(x)]^n\,f'(x)\,dx = \dfrac{[f(x)]^{n+1}}{n+1} + C$ for $n \neq -1$ (and the $n = -1$ case is the log pattern). The shape $\displaystyle\int \dfrac{f'(x)}{[f(x)]^2}\,dx = -\dfrac{1}{f(x)} + C$ is the $n = -2$ instance.

Power-of-a-function rule

\int [f(x)]^n\,f'(x)\,dx = \dfrac{[f(x)]^{n+1}}{n+1} + C \quad (n \neq -1)

Worked example

Evaluate

\displaystyle\int (x^3 + 2x + 1)^4\,(3x^2 + 2)\,dx

Let $u = x^3 + 2x + 1$ . Then $du = (3x^2 + 2)\,dx$ — exactly the second factor.
Rewrite: $\int u^4\,du = \dfrac{u^5}{5} + C$ .
Back-substitute.

Answer:

\dfrac{(x^3 + 2x + 1)^5}{5} + C

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 3Indefinite IntegrationMODERATE

\int\frac{2+\cos x}{(2x+\sin x)^2}\,dx=

[Q103 · 13th May Shift 2 · 2024]

n = −1 is NOT this rule

If the power is

-1

(i.e.

\int f'/f

), the power rule blows up. That case is the log pattern. Every other integer/fraction power uses

\dfrac{u^{n+1}}{n+1}

Drill 1 more on power of a function times its derivative

Concept 4 of 6

Root and Linear-Radical Substitutions

Intuition

A square root of a linear (or simple) expression is cleared by substituting the root itself, or the inside of the root, as

u

. The substitution removes the radical and leaves a polynomial or standard form.

Definition

For integrals containing $\sqrt{ax+b}$ , substitute $t = \sqrt{ax+b}$ (so $x$ and $dx$ are expressed via $t$ ), or substitute $u = ax+b$ . For $\sqrt{x}$ shapes, $t = \sqrt{x}$ gives $x = t^2,\ dx = 2t\,dt$ . The radical disappears and the integrand becomes rational/polynomial in $t$ .

Linear-radical substitution

t = \sqrt{ax+b}\ \Rightarrow\ x = \dfrac{t^2 - b}{a},\quad dx = \dfrac{2t}{a}\,dt

Worked example

Evaluate

\displaystyle\int \dfrac{1}{1 + \sqrt{x}}\,dx

Let $t = \sqrt{x}$ , so $x = t^2$ and $dx = 2t\,dt$ .
Rewrite: $\int \dfrac{1}{1+t}\,(2t\,dt) = 2\int \dfrac{t}{1+t}\,dt = 2\int\!\left(1 - \dfrac{1}{1+t}\right)dt$ .
Integrate: $2\big(t - \log|1+t|\big) + C$ .
Back-substitute $t = \sqrt{x}$ .

Answer:

2\sqrt{x} - 2\log\!\left|1 + \sqrt{x}\right| + C

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 4Indefinite IntegrationMODERATE

\int\frac{\sqrt{x}}{x+1}\,dx=

[Q131 · 10th May Shift 1 · 2023]

Re-express EVERY x, including dx

After

t = \sqrt{x}

, both the integrand AND

dx = 2t\,dt

must be rewritten. Leaving a stray

x

or the old

dx

behind is the classic substitution error.

Drill 10 more on root and linear-radical substitutions

Concept 5 of 6

Reciprocal and Take-out-the-Power Substitutions

Intuition

When

x

is trapped inside high powers or a root, divide through by a power of

x

(or factor the dominant power out of the root). What's left is a reciprocal combination like

1+x^{-4}

x\pm\dfrac{k}{x}

whose derivative is sitting right there in the numerator — substitute it and the mess collapses. This is the chapter's single most-repeated HARD substitution.

Definition

Two faces of the same idea:

Take-out-the-power: for $\displaystyle\int \dfrac{dx}{x^2(x^4+1)^{3/4}}$ -type integrals, pull the dominant power out of the root — $(x^4+1)^{3/4}=x^3(1+x^{-4})^{3/4}$ — then substitute $u = 1 + x^{-4}$ .
**Reciprocal $t = x \pm \dfrac{k}{x}$ **: for a denominator quadratic in $x^2$ with a matching numerator, divide top and bottom by $x^2$ . Then $\dfrac{d}{dx}\!\left(x\mp\dfrac{k}{x}\right) = 1\pm\dfrac{k}{x^2}$ is exactly the new numerator, and the integral becomes a standard $t^2+1$ arctan (or a clean power).

The reciprocal substitution

t = x \pm \dfrac{k}{x} \;\Rightarrow\; dt = \left(1 \mp \dfrac{k}{x^2}\right)dx

\pm k/xsign chosen so $dt$ matches the numerator after dividing by $x^2$

Worked example

Evaluate

\displaystyle\int \dfrac{dx}{x^2(x^4+1)^{3/4}}

Take the dominant power out of the root: $(x^4+1)^{3/4} = x^3\,(1+x^{-4})^{3/4}$ .
Integrand $= \dfrac{1}{x^2\cdot x^3(1+x^{-4})^{3/4}} = x^{-5}\,(1+x^{-4})^{-3/4}$ .
Let $u = 1 + x^{-4}$ , so $du = -4x^{-5}\,dx$ , i.e. $x^{-5}\,dx = -\tfrac14\,du$ .
Integrate: $-\tfrac14\!\int u^{-3/4}\,du = -\tfrac14\cdot 4u^{1/4} = -u^{1/4}$ . Back-substitute.

Answer:

-\dfrac{(x^4+1)^{1/4}}{x} + C

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 5Indefinite 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 · Shift 1 · 2023]

Pick the sign of t = x ± k/x from the numerator

After dividing by

x^2

, if the numerator is

1 - \dfrac{k}{x^2}

use

t = x + \dfrac{k}{x}

; if it is

1 + \dfrac{k}{x^2}

use

t = x - \dfrac{k}{x}

. The substitution only works when

dt

reproduces the numerator exactly.

Drill 6 more on reciprocal and take-out-the-power substitutions

Concept 6 of 6

Exponential and Special Substitutions

Intuition

When an exponential is buried inside a root or a tower of exponents, substitute the exponential itself.

t = e^x

(so

dt = e^x dx

) clears most

e^x

integrals; nested towers like

3^{3^x}

substitute the inner power.

Definition

For integrands built from $e^x$ : substitute $t = e^x$ , giving $dt = e^x\,dx$ ; a stray $e^x$ in the numerator becomes $dt$ and the rest becomes rational in $t$ . For exponential towers $\int a^{a^x}\,a^x\,dx$ , substitute $u = a^x$ (then $du = a^x \log a\,dx$ ).

Exponential substitution

t = e^{x}\ \Rightarrow\ dt = e^{x}\,dx

Worked example

Evaluate

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

Let $t = e^x$ , so $dt = e^x\,dx$ . Note $e^{2x} = (e^x)^2 = t^2$ .
Rewrite: $\int \dfrac{dt}{1 + t^2} = \tan^{-1}t + C$ .
Back-substitute $t = e^x$ .

Answer:

\tan^{-1}(e^x) + C

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 6Indefinite IntegrationMODERATE

\int\sqrt{e^x - 1}\,dx =

[Q119 · 9th May Shift 2 · 2023]

Towers: substitute the INNER exponential

For

\int a^{a^x} a^x\,dx

, the right substitution is

u = a^x

, not

u = a^{a^x}

. The stray

a^x\,dx

is what becomes

du

(up to

\log a

Drill 6 more on exponential and special substitutions

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 (6)

The Substitution Rule
Substitution rule
$\int f(g(x))\,g'(x)\,dx = \int f(u)\,du,\quad u = g(x)$
The f'(x)/f(x) → log Pattern
Logarithmic integral
$\int \dfrac{f'(x)}{f(x)}\,dx = \log|f(x)| + C$
Power of a Function times its Derivative
Power-of-a-function rule
$\int [f(x)]^n\,f'(x)\,dx = \dfrac{[f(x)]^{n+1}}{n+1} + C \quad (n \neq -1)$
Root and Linear-Radical Substitutions
Linear-radical substitution
$t = \sqrt{ax+b}\ \Rightarrow\ x = \dfrac{t^2 - b}{a},\quad dx = \dfrac{2t}{a}\,dt$
Reciprocal and Take-out-the-Power Substitutions
The reciprocal substitution
$t = x \pm \dfrac{k}{x} \;\Rightarrow\; dt = \left(1 \mp \dfrac{k}{x^2}\right)dx$
Exponential and Special Substitutions
Exponential substitution
$t = e^{x}\ \Rightarrow\ dt = e^{x}\,dx$

Watch out for (6)

Adjust for the missing constant
→ The Substitution Rule
Engineer the numerator into f'(x) + leftover
→ The f'(x)/f(x) → log Pattern
n = −1 is NOT this rule
→ Power of a Function times its Derivative
Re-express EVERY x, including dx
→ Root and Linear-Radical Substitutions
Pick the sign of t = x ± k/x from the numerator
→ Reciprocal and Take-out-the-Power Substitutions
Towers: substitute the INNER exponential
→ Exponential and Special Substitutions

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