MHT-CET Maths · Indefinite Integration

Integration by Parts

Integrate a product by trading it for an easier integral — choose u by LIATE, and watch for the cyclic and eˣ[f+f'] shortcuts.

All Indefinite Integration notes MHT-CET Maths strategy

Why this matters

23 PYQs, and the chapter's second-hardest pocket (15 of 23 are HARD). Three patterns dominate: the LIATE choice for ordinary products; the cyclic integrals (∫eˣ sin x, ∫sin(log x)) that return to themselves; and the recurring eˣ[f(x)+f'(x)] → eˣ f(x) family that MHT-CET tests almost every year. Recognising the eˣ[f+f'] shape on sight turns a HARD question into a one-line answer.

Concept 1 of 5

Integration by Parts and the LIATE Rule

Intuition

To integrate a product, call one factor

u

(to differentiate) and the other

dv

(to integrate). Pick

u

by LIATE — Log, Inverse-trig, Algebraic, Trig, Exponential — so that differentiating

u

makes the problem simpler.

Definition

Integration by parts: $\displaystyle\int u\,dv = uv - \int v\,du$ . Choose $u$ as the function that appears EARLIEST in LIATE (Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential) — it differentiates toward something simpler, while the rest is $dv$ . A lone $\log x$ or $\tan^{-1}x$ is integrated by taking $dv = dx$ .

Integration by parts

\int u\,dv = uv - \int v\,du

ufactor to differentiate (earliest in LIATE)
dvfactor to integrate (the rest, including $dx$ )

Worked example

Evaluate

\displaystyle\int \log x\,dx

There is only one function — take $u = \log x$ and $dv = dx$ (Log is first in LIATE).
Then $du = \dfrac{1}{x}\,dx$ and $v = x$ .
Apply parts: $uv - \int v\,du = x\log x - \int x\cdot\dfrac{1}{x}\,dx = x\log x - \int 1\,dx$ .
Finish: $x\log x - x + C$ .

Answer:

x\log x - x + C

Practice this conceptself-check · 4 quick reps

Try it yourself

Evaluate

\displaystyle\int x^2 e^x\,dx

Practice — Level 1 (4 reps)

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

1.
In $\int x e^x\,dx$ , what is $u$ by LIATE?
2.
$\int x e^x\,dx$
3.
$\int \tan^{-1}x\,dx$
4.
$\int x\cos x\,dx$

From the bank · past-year question

Example 1Indefinite IntegrationHARD

\int\sin^{-1}\!\left(\frac{2x}{1+x^2}\right)dx =

[Q146 · 10th May Shift 2 · 2023]

A lone log or inverse-trig still uses parts

\int \log x\,dx

and

\int \tan^{-1}x\,dx

look like single functions, but they are integrated by parts with

dv = dx,\ v = x

. There is no direct formula.

Drill 5 more on integration by parts and the liate rule

Concept 2 of 5

Cyclic Integrals (Return-to-Self)

Intuition

For

\int e^{ax}\sin bx\,dx

and

\int \sin(\log x)\,dx

, applying parts twice brings back the ORIGINAL integral. Move it to the left side and solve algebraically — no third round needed.

Definition

Apply integration by parts twice. The same integral $I$ reappears on the right with a coefficient; collect it: $I = (\text{boundary terms}) + kI\Rightarrow I(1-k) = \text{terms}$ . For $\int \sin(\log x)\,dx$ , the substitution $x = e^t$ turns it into $\int e^t \sin t\,dt$ , the classic cyclic form.

The cyclic result

\int e^{ax}\sin bx\,dx = \dfrac{e^{ax}(a\sin bx - b\cos bx)}{a^2 + b^2} + C

Worked example

Evaluate

\displaystyle\int e^{2x} \sin 3x\,dx

This is a cyclic integral — applying parts twice returns the original integral, so use the cyclic formula directly.
$\int e^{ax}\sin bx\,dx = \dfrac{e^{ax}(a\sin bx - b\cos bx)}{a^2 + b^2}$ .
Substitute $a = 2,\ b = 3$ : $\dfrac{e^{2x}(2\sin 3x - 3\cos 3x)}{4 + 9}$ .

Answer:

\dfrac{e^{2x}(2\sin 3x - 3\cos 3x)}{13} + C

Practice this conceptself-check · 4 quick reps

Try it yourself

Evaluate

\displaystyle\int \cos(\log x)\,dx

Practice — Level 1 (4 reps)

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

1.
$\int e^x \sin x\,dx$
2.
$\int e^x \cos x\,dx$
3.
Substitution for $\int \sin(\log x)\,dx$ ?
4.
After two by-parts rounds, the original integral I is solved by?

From the bank · past-year question

Example 2Indefinite IntegrationMODERATE

I=\int\sin(\log x)\,dx

, then I is given by

[Q137 · 11th May Shift 2 · 2024]

Stop after two rounds — don't loop forever

The point of a cyclic integral is that the original

I

reappears after two rounds. Recognise it and solve algebraically; a third by-parts just sends you in circles.

Drill 2 more on cyclic integrals (return-to-self)

Concept 3 of 5

The eˣ[f(x) + f'(x)] Family

Intuition

Whenever an integrand is

e^x

times 'some function plus its own derivative', the answer is just

e^x

times that function. Spotting the

f + f'

pattern collapses a scary integral to one line.

Definition

$\displaystyle\int e^x\big[f(x) + f'(x)\big]\,dx = e^x f(x) + C$ . The work is REWRITING the integrand into this shape — using identities so that one part is a function $f$ and the rest is exactly its derivative $f'$ . A close cousin: $\displaystyle\int e^x\dfrac{1 + x\log x}{x}$ -style problems all reduce to spotting $f + f'$ .

The eˣ[f + f'] shortcut

\int e^x\big[f(x) + f'(x)\big]\,dx = e^x f(x) + C

f(x)the function whose value lands in the answer
f'(x)its derivative — must be the other half of the bracket

Worked example

Evaluate

\displaystyle\int e^x\!\left(\log x + \dfrac{1}{x}\right)dx

The bracket is $f(x) + f'(x)$ with $f(x) = \log x$ , since $f'(x) = \dfrac{1}{x}$ .
Apply the shortcut $\int e^x[f + f']\,dx = e^x f(x)$ .
So the integral is $e^x \log x + C$ .

Answer:

e^x \log x + C

Practice this conceptself-check · 4 quick reps

Try it yourself

Evaluate

\displaystyle\int e^x\!\left(\dfrac{x - 1}{x^2}\right)dx

Practice — Level 1 (4 reps)

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

1.
$\int e^x(\sin x + \cos x)\,dx$
2.
$\int e^x\!\left(\tan x + \sec^2 x\right)dx$
3.
$\int e^x\!\left(\dfrac1x - \dfrac{1}{x^2}\right)dx$
4.
$\int e^x(1 + x)\,dx$

From the bank · past-year question

Example 3Indefinite IntegrationMODERATE

\displaystyle\int e^x(1-\cot x+\cot^2 x)\,dx =

[Q125 · 10th May Shift 2 · 2024]

Use identities to expose f + f'

The bracket rarely arrives as a clean

f + f'

. Apply identities first — e.g.

1 + \cot^2 = \csc^2

— so that one term is a function and the other is its exact derivative. Then the answer is immediate.

Drill 13 more on the eˣ[f(x) + f'(x)] family

Concept 4 of 5

Integrals of √(quadratic) — Standard Results (syllabus reference)

Intuition

These three results — proved by integration by parts in the textbook — give the integral of a square root of a quadratic. Recent MHT-CET papers in this bank have not tested them directly, but they are on the syllabus and a question can reduce to one after completing the square, so keep them on hand.

Definition

The three standard results (each derived by taking $\sqrt{\,\cdot\,}$ as the first function and $1$ as the second):

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

For a general $\sqrt{ax^2+bx+c}$ , complete the square first, then match one of these three.

Square root of a quadratic — the three results

\int \sqrt{a^2 - x^2}\,dx = \dfrac{x}{2}\sqrt{a^2 - x^2} + \dfrac{a^2}{2}\sin^{-1}\dfrac{x}{a} + C

a^2 - x^2arcsin result
a^2 + x^2,\ x^2 - a^2log results (signs differ)

Worked example

Evaluate

\displaystyle\int \sqrt{9 - x^2}\,dx

Match the first result with $a^2 = 9\Rightarrow a = 3$ .
$\displaystyle\int \sqrt{9 - x^2}\,dx = \dfrac{x}{2}\sqrt{9 - x^2} + \dfrac{9}{2}\sin^{-1}\dfrac{x}{3} + C$ .

Answer:

\dfrac{x}{2}\sqrt{9 - x^2} + \dfrac{9}{2}\sin^{-1}\dfrac{x}{3} + C

Practice this concept4 quick reps

Practice — Level 1 (4 reps)

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

1.
$\int \sqrt{a^2-x^2}\,dx$ uses which inverse-trig term?
2.
$\int \sqrt{x^2-a^2}\,dx$ — sign before the log term?
3.
$\int \sqrt{4-x^2}\,dx$
4.
For $\int\sqrt{x^2+16}\,dx$ , the constant before log is?

Syllabus result, not a current bank pattern

This bank's recent CET questions keep the square root in a DENOMINATOR (handled by the numerator-split + completing-the-square concepts). The

\int\sqrt{\text{quadratic}}\,dx

form here is on the syllabus and could appear — but don't expect it among the tagged drills, because the live bank has none yet.

Concept 5 of 5

Generalised (Tabular) By-Parts — a shortcut

Intuition

When the first function is a polynomial, repeated by-parts produces an alternating pattern that you can write in one line: differentiate the polynomial down to zero, integrate the other factor repeatedly, and alternate signs. It is the same answer as ordinary by-parts, faster.

Definition

If $u$ is a polynomial (so some derivative $u^{(n)} = 0$ ) and $v$ the second function, then $\displaystyle\int u\,v\,dx = u\,v_1 - u'\,v_2 + u''\,v_3 - u'''\,v_4 + \cdots$ , where a prime is a derivative of $u$ and a subscript $k$ means integrate $v$ $k$ times. The series terminates because the polynomial's derivatives reach zero.

Tabular by-parts series

\int u\,v\,dx = u\,v_1 - u'\,v_2 + u''\,v_3 - u'''\,v_4 + \cdots

u', u'', \ldotssuccessive derivatives of the polynomial $u$
v_1, v_2, \ldotssuccessive integrals of $v$

Worked example

Evaluate

\displaystyle\int x^2 \cos 3x\,dx

by the tabular method.

Derivatives of $u = x^2$ : $x^2,\ 2x,\ 2,\ 0$ . Integrals of $\cos 3x$ : $\tfrac13\sin 3x,\ -\tfrac19\cos 3x,\ -\tfrac{1}{27}\sin 3x$ .
Alternate signs $+,-,+$ : $x^2\cdot\tfrac13\sin 3x - 2x\cdot(-\tfrac19\cos 3x) + 2\cdot(-\tfrac{1}{27}\sin 3x)$ .
Simplify.

Answer:

\dfrac{1}{3}x^2\sin 3x + \dfrac{2}{9}x\cos 3x - \dfrac{2}{27}\sin 3x + C

Practice this concept4 quick reps

Practice — Level 1 (4 reps)

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

1.
Tabular by-parts terminates when?
2.
Sign pattern across the terms?
3.
$\int x e^x\,dx$ by table: derivatives $x,1,0$ ; integrals $e^x, e^x$ .
4.
Which factor should be $u$ (the one you differentiate)?

Only for a polynomial first function

The shortcut relies on the polynomial's derivatives reaching zero. If neither factor is a polynomial (e.g.

\int e^x\sin x\,dx

), the table never terminates — use ordinary or cyclic by-parts instead.

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 (5)

Integration by Parts and the LIATE Rule
Integration by parts
$\int u\,dv = uv - \int v\,du$
Cyclic Integrals (Return-to-Self)
The cyclic result
$\int e^{ax}\sin bx\,dx = \dfrac{e^{ax}(a\sin bx - b\cos bx)}{a^2 + b^2} + C$
The eˣ[f(x) + f'(x)] Family
The eˣ[f + f'] shortcut
$\int e^x\big[f(x) + f'(x)\big]\,dx = e^x f(x) + C$
Integrals of √(quadratic) — Standard Results (syllabus reference)
Square root of a quadratic — the three results
$\int \sqrt{a^2 - x^2}\,dx = \dfrac{x}{2}\sqrt{a^2 - x^2} + \dfrac{a^2}{2}\sin^{-1}\dfrac{x}{a} + C$
Generalised (Tabular) By-Parts — a shortcut
Tabular by-parts series
$\int u\,v\,dx = u\,v_1 - u'\,v_2 + u''\,v_3 - u'''\,v_4 + \cdots$

Watch out for (5)

A lone log or inverse-trig still uses parts
→ Integration by Parts and the LIATE Rule
Stop after two rounds — don't loop forever
→ Cyclic Integrals (Return-to-Self)
Use identities to expose f + f'
→ The eˣ[f(x) + f'(x)] Family
Syllabus result, not a current bank pattern
→ Integrals of √(quadratic) — Standard Results (syllabus reference)
Only for a polynomial first function
→ Generalised (Tabular) By-Parts — a shortcut

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\sin\!\sqrt{x}\,dx

is equal to

[Q121 · 3rd May Shift 2 · 2023]

Example 2Indefinite IntegrationHARD

\int\cos(\log x)\,dx =

[Q128 · 11th May Shift 2 · 2023]

Example 3Indefinite IntegrationHARD

\int\!\left(1+x-\frac{1}{x}\right)e^{x+\frac{1}{x}}\,dx

equals

[Q108 · 4th May Shift 1 · 2023]

Example 4Indefinite IntegrationMODERATE

\int e^{x^2}\cdot x^3\,dx = e^{x^2}f(x)+c

and

f(1)=0

(where

c

is a constant of integration), then the value of

f(x)

[Q131 · 15th May Shift 2 · 2023]

Example 5Indefinite IntegrationHARD

\int\frac{1+\sin(\log x)}{1+\cos(\log x)}\,dx=

[Q133 · 10th May Shift 1 · 2023]

Drill every past-year question on this subtopic

23 questions from the bank — paginated, with cart and Word-export support.

Drill the 23 questions

Related notes

Substitution — the fallback when no product structure helps →