MathsISC Class 12 · Integrals

Integration by Parts: Formula, ILATE Rule and Special Forms

Q: How do I decide which function is first in integration by parts?

Use ILATE: inverse trigonometric, logarithmic, algebraic, trigonometric, exponential. The function whose type comes first in this list is the first function u, because it becomes simpler when differentiated.

Q: What if there is only one function, like log x?

Treat it as a product with 1. Take log x as the first function and 1 as the second; the integral of 1 is x, which cancels the 1/x from the derivative of log x.

Q: What is a cyclic integral?

It is an integral such as ∫eˣ sin x dx where applying parts twice brings back the original integral. Call it I, move the I terms to one side and solve for I.

Q: When can I use ∫eˣ[f(x) + f′(x)] dx = eˣf(x) + C?

Whenever the integrand is eˣ multiplied by a function plus its own derivative. Rewrite the bracket to check, for example x/(1 + x)² = 1/(1 + x) − 1/(1 + x)².

Q: How many marks does integration by parts carry in ISC?

It appears regularly as a 4-mark question (for example ∫x² cos x dx in 2024 and ∫(log x)² dx in 2025) and in 1-mark items in Section A. Calculus as a whole carries 35 of the 80 marks in the 2027 paper.

Updated 15 October 2026 · Fact-checked

Integration by parts integrates a product: ∫u v dx = u∫v dx − ∫(u′ ∫v dx) dx. Choose u, the first function, as the one that becomes simpler when differentiated, using the order ILATE: inverse trigonometric, logarithmic, algebraic, trigonometric, exponential. Repeat if needed, and watch for cyclic integrals.

For a product, take as u whichever type appears highest in ILATE; the other factor becomes v, which you integrate.

Illustration drawn for QuestionBuddy.

Understand Integration by Parts

Where the formula comes from. The product rule says $\frac{d}{d x} (u w) = u \frac{d w}{d x} + w \frac{d u}{d x}$ . Integrating and rearranging, with $w = \int v d x$ (so $\frac{d w}{d x} = v$ ), gives
$\int u v d x = u \int v d x - \int (\frac{d u}{d x} \int v d x) d x .$
In words: first times the integral of the second, minus the integral of (derivative of the first times the integral of the second). It turns an integral of a product into another integral, which should be easier.

Choosing the first function. The method works only if the new integral is simpler. That happens when $u$ gets simpler on differentiating and $v$ is easy to integrate. The ISC 2024 and 2025 examiners both reported wrong choices of first and second functions. Use the order ILATE to pick $u$ , taking whichever comes first:

Inverse trigonometric ( $tan^{- 1} x$ , $sin^{- 1} x$ )
Logarithmic ( $lo g x$ )
Algebraic ( $x$ , $x^{2}$ )
Trigonometric ( $sin x$ , $cos x$ )
Exponential ( $e^{x}$ )

So in $\int x cos x d x$ take $u = x$ ; in $\int x lo g x d x$ take $u = lo g x$ .

A single function. To integrate $lo g x$ , $(lo g x)^{2}$ , $tan^{- 1} x$ or $sin^{- 1} x$ alone, write it as a product with 1 and take 1 as the second function: $\int lo g x d x = x lo g x - \int x \cdot \frac{1}{x} d x = x lo g x - x + C$ . ISC 2025 asked $\int (lo g x)^{2} d x$ , which needs this idea twice.

Repeated parts. For $\int x^{2} cos x d x$ (ISC 2024) or $\int x^{2} e^{x} d x$ , one application lowers the power of $x$ by one; apply the method again until the algebraic factor disappears. Keep track of signs by bracketing each new integral.

Cyclic integrals. In $\int e^{a x} sin b x d x$ , neither factor gets simpler. Apply parts twice (keeping the same type of function as $u$ both times); the original integral reappears on the right. Call it $I$ and solve the equation for $I$ :
$\int e^{a x} sin b x d x = \frac{e ^{a x}}{a ^{2} + b ^{2}} (a sin b x - b cos b x) + C .$

**The $e^{x} [f (x) + f^{'} (x)]$ rule.** If the integrand is $e^{x}$ times a function plus its own derivative, then
$\int e^{x} [f (x) + f^{'} (x)] d x = e^{x} f (x) + C .$
Spot it by writing the bracket in the form $f + f^{'}$ . For example, $\frac{x e ^{x}}{( 1 + x ) ^{2}} = e^{x} [\frac{1}{1 + x} - \frac{1}{( 1 + x ) ^{2}}]$ , with $f (x) = \frac{1}{1 + x}$ . The ISC 2024 MCQ $\int (cot x - cosec^{2} x) e^{x} d x$ is this rule with $f (x) = cot x$ .

Standard results from parts (listed in the syllabus):
$\int a^{2} - x^{2} d x = \frac{x}{2} a^{2} - x^{2} + \frac{a ^{2}}{2} sin^{- 1} \frac{x}{a} + C$ , and similar forms for $x^{2} \pm a^{2}$ .

Definite integrals. With limits, $\int_{a}^{b} uv d x = [u \int v d x]_{a}^{b} - \int_{a}^{b} u^{'} (\int v d x) d x$ . You can also find the antiderivative first and substitute the limits at the end.

Integration by Parts: formulas to remember

Integration by parts: $\int u v d x = u \int v d x - \int (u^{'} \int v d x) d x$; u is the first function (ILATE).
ILATE order: $Inverse trig \to Log \to Algebraic \to Trig \to Exponential$; Choose u as the type that comes first.
Log x: $\int lo g x d x = x lo g x - x + C$; Take 1 as the second function.
Inverse tangent: $\int tan^{- 1} x d x = x tan^{- 1} x - \frac{1}{2} lo g (1 + x^{2}) + C$; Same method for sin⁻¹x: x sin⁻¹x + √(1 − x²) + C.
eˣ rule: $\int e^{x} [f (x) + f^{'} (x)] d x = e^{x} f (x) + C$; Rewrite the bracket as f + f′.
Cyclic: $\int e^{a x} sin b x d x = \frac{e ^{a x} ( a s i n b x - b c o s b x )}{a ^{2} + b ^{2}} + C$; cos version: e^{ax}(a cos bx + b sin bx)/(a² + b²).
Square-root form: $\int a^{2} - x^{2} d x = \frac{x}{2} a^{2} - x^{2} + \frac{a ^{2}}{2} sin^{- 1} \frac{x}{a} + C$; Derived by parts.

How to solve Integration by Parts questions

Use these steps for every integration-by-parts question.

1Check first whether the integrand is $e^{x} [f (x) + f^{'} (x)]$ ; if so, write the answer $e^{x} f (x) + C$ at once.
2Identify the two factors and choose the first function $u$ by ILATE. For a single function such as $lo g x$ , take $u$ as that function and $v = 1$ .
3Write the formula with your $u$ and $v$ before substituting, so that the structure is visible.
4Integrate $v$ , differentiate $u$ , and simplify the new integral.
5If the new integral is still a product with a polynomial factor, apply parts again.
6If the original integral reappears (cyclic case), call it $I$ , collect the $I$ terms and solve for $I$ .
7Add $+ C$ for an indefinite integral; for a definite integral substitute the limits and simplify.

Integration by Parts in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Integration by parts for total energy

Suppose a heater, once switched on, draws power $P (t) = 2 t e^{- t}$ kilowatts, where $t$ is the time in hours, rising quickly and then tailing off.

Find the total energy used in the first 3 hours, $E = \int_{0}^{3} 2 t e^{- t} d t$ kWh.

Try it first, then show the solutionSolution

By parts with $u = t$ , $v = e^{- t}$ : $\int t e^{- t} d t = - t e^{- t} - \int (- e^{- t}) d t = - t e^{- t} - e^{- t} = - e^{- t} (t + 1)$ .
$E = 2 [- e^{- t} (t + 1)]_{0}^{3} = 2 [- 4 e^{- 3} + 1] = 2 (1 - 4 e^{- 3})$ .
$e^{- 3} \approx 0.0498$ , so $E \approx 2 (1 - 0.199) \approx 1.60$ kWh.

Answer: About 1.60 kWh, exactly $2 (1 - 4 e^{- 3})$ kWh.

Memory hook: Rising times fading: t·e⁻ᵗ needs parts, with t as the first function.

Integration by parts with a trigonometric factor

Imagine a tank that drains at a rate of $t sin t$ hundred litres per minute during the first $π$ minutes after a valve is opened (the rate starts at zero, rises, then falls back to zero).

How much water drains in the first $π$ minutes?

Try it first, then show the solutionSolution

By parts with $u = t$ , $v = sin t$ : $\int t sin t d t = - t cos t + \int cos t d t = - t cos t + sin t$ .
$[- t cos t + sin t]_{0}^{π} = (- π cos π + 0) - 0 = π$ .
So $π$ hundred litres, about 314 litres, drain.

Answer: About 314 litres ( $100 π$ litres).

Memory hook: Algebraic before trigonometric: in t sin t, t is the first function.

Tricks to remember Integration by Parts

Short memory aids. Say each one aloud once, then use it on a question the same day.

ILATE
Inverse trig, Log, Algebraic, Trig, Exponential: the earlier type is the first function.
Use it when: Choosing u correctly every time.
First × ∫second − ∫(first′ × ∫second)
Say it as a rhythm: "first, integral second; minus integral of derivative-first, integral second".
Use it when: Writing the formula without hesitation.
Lonely log
A single log or inverse trig function? Multiply by 1 and let 1 be the second function.
Use it when: ∫log x dx, ∫(log x)² dx, ∫tan⁻¹x dx.
Spot f + f′
eˣ times (function + its derivative) integrates to eˣ times the function.
Use it when: One-line answers to eˣ[…] integrals.

Integration by Parts: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: A basic product · 1 mark · MCQBoard level

$\int x e^{x} d x$ equals:
(a) $x e^{x} + C$
(b) $e^{x} (x - 1) + C$
(c) $e^{x} (x + 1) + C$
(d) $\frac{x ^{2}}{2} e^{x} + C$

Try it first, then show the solutionSolution

$u = x$ , $v = e^{x}$ : $x e^{x} - \int e^{x} d x = x e^{x} - e^{x} + C$ .

Answer: (b) $e^{x} (x - 1) + C$ .

Example 2: Algebraic times trigonometric · 2 marks · Short answerBoard level

Evaluate $\int x cos x d x$ .

Try it first, then show the solutionSolution

By ILATE, $u = x$ (algebraic), $v = cos x$ . (½ mark)
$\int x cos x d x = x sin x - \int 1 \cdot sin x d x$ . (1 mark)
$= x sin x + cos x + C$ . (½ mark)

Answer: $x sin x + cos x + C$ .

Example 3: Parts twice · 3 marks · Short answerHarder board level

Evaluate $\int x^{2} e^{2 x} d x$ .

Try it first, then show the solutionSolution

$u = x^{2}$ , $v = e^{2 x}$ : $\int x^{2} e^{2 x} d x = \frac{x ^{2} e ^{2 x}}{2} - \int 2 x \cdot \frac{e ^{2 x}}{2} d x = \frac{x ^{2} e ^{2 x}}{2} - \int x e^{2 x} d x$ . (1 mark)
Again with $u = x$ : $\int x e^{2 x} d x = \frac{x e ^{2 x}}{2} - \int \frac{e ^{2 x}}{2} d x = \frac{x e ^{2 x}}{2} - \frac{e ^{2 x}}{4}$ . (1 mark)
So $\int x^{2} e^{2 x} d x = \frac{x ^{2} e ^{2 x}}{2} - \frac{x e ^{2 x}}{2} + \frac{e ^{2 x}}{4} + C = \frac{e ^{2 x}}{4} (2 x^{2} - 2 x + 1) + C$ . (1 mark)

Answer: $\frac{e ^{2 x}}{4} (2 x^{2} - 2 x + 1) + C$ . (ISC 2024 asked $\int x^{2} cos x d x$ , which needs the same two steps.)

Example 4: Spotting eˣ[f + f′] · 2 marks · Short answerHarder board level

Evaluate $\int e^{x} (tan x + lo g sec x) d x$ , on an interval where $sec x > 0$ .

Try it first, then show the solutionSolution

With $f (x) = lo g sec x$ , $f^{'} (x) = \frac{s e c x t a n x}{s e c x} = tan x$ , so the integrand is $e^{x} [f (x) + f^{'} (x)]$ . (1 mark)
Hence the integral is $e^{x} f (x) + C = e^{x} lo g sec x + C$ . (1 mark)

Answer: $e^{x} lo g sec x + C$ .

Example 5: An inverse function on its own · 3 marks · Short answerHarder board level

Evaluate $\int tan^{- 1} x d x$ .

Try it first, then show the solutionSolution

Write as $\int tan^{- 1} x \cdot 1 d x$ with $u = tan^{- 1} x$ , $v = 1$ . (½ mark)
$= x tan^{- 1} x - \int \frac{x}{1 + x ^{2}} d x$ . (1 mark)
$\int \frac{x}{1 + x ^{2}} d x = \frac{1}{2} lo g (1 + x^{2})$ (substitution). (1 mark)
So the integral is $x tan^{- 1} x - \frac{1}{2} lo g (1 + x^{2}) + C$ . (½ mark)

Answer: $x tan^{- 1} x - \frac{1}{2} lo g (1 + x^{2}) + C$ .

Example 6: A cyclic integral with limits · 5 marks · Long answerHOTS / competency

Evaluate $\int e^{2 x} sin x d x$ and hence $\int_{0}^{π} e^{2 x} sin x d x$ .

Try it first, then show the solutionSolution

Let $I = \int e^{2 x} sin x d x$ . Take $u = sin x$ , $v = e^{2 x}$ : $I = \frac{e ^{2 x}}{2} sin x - \frac{1}{2} \int e^{2 x} cos x d x$ . (1 mark)
Again with $u = cos x$ : $\int e^{2 x} cos x d x = \frac{e ^{2 x}}{2} cos x + \frac{1}{2} \int e^{2 x} sin x d x = \frac{e ^{2 x}}{2} cos x + \frac{I}{2}$ . (1 mark)
So $I = \frac{e ^{2 x}}{2} sin x - \frac{e ^{2 x}}{4} cos x - \frac{I}{4}$ , giving $\frac{5 I}{4} = \frac{e ^{2 x}}{4} (2 sin x - cos x)$ . (1 mark)
$I = \frac{e ^{2 x}}{5} (2 sin x - cos x) + C$ . (1 mark)
Definite: $[\frac{e ^{2 x}}{5} (2 sin x - cos x)]_{0}^{π} = \frac{e ^{2 π} ( 0 + 1 )}{5} - \frac{1 \cdot ( 0 - 1 )}{5} = \frac{e ^{2 π} + 1}{5}$ . (1 mark)

Answer: $\frac{e ^{2 x}}{5} (2 sin x - cos x) + C$ ; the definite integral is $\frac{e ^{2 π} + 1}{5}$ .

Where students lose marks in Integration by Parts

The slips examiners see most often, and what to write instead.

Mistake: Choosing the wrong first function
Both the 2024 and 2025 examiners reported wrong choices of first and second functions, which make the new integral harder.
Wrong working
$\int x cos x d x$ with $u = cos x$ : $\frac{x ^{2}}{2} cos x + \int \frac{x ^{2}}{2} sin x d x$ (worse).
Right working
With $u = x$ : $x sin x - \int sin x d x = x sin x + cos x + C$ .
Do this: Apply ILATE before writing anything.
Usually costs: Most of the question.
Mistake: Not writing the standard form
The 2025 examiners noted candidates who did not write the formula, and then lost track of signs.
Do this: Write the formula with your u and v substituted as the first line.
Usually costs: Method marks.
Mistake: Sign errors in repeated parts
The minus sign in front of the new integral must multiply everything that comes from it.
Do this: Put each new integral in brackets: x²eˣ − [2xeˣ − 2eˣ].
Usually costs: 1 mark.
Mistake: Treating a cyclic integral as endless
Students keep applying parts and never finish.
Do this: After two applications, look for the original integral, call it I, and solve for I.
Usually costs: The whole question.

Exam tips

State your choice: "taking $x^{2}$ as the first function and $cos x$ as the second". Examiners give credit for the correct choice.
Keep $\int v d x$ without a constant in the middle of the working; add one $+ C$ at the end.
For $e^{x} [\dots]$ integrals, always test the $f + f^{'}$ form first; it can save five lines.
In cyclic integrals, keep the same type of function (trigonometric) as $u$ in both applications, or the original integral will not return.
With limits, evaluate the boundary term $[u \int v]_{a}^{b}$ carefully; it is often zero at one limit.

Practice questions on Integration by Parts

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
$\int x lo g x d x$ equals:
(a) $\frac{x ^{2}}{2} lo g x + C$
(b) $\frac{x ^{2}}{2} lo g x - \frac{x ^{2}}{4} + C$
(c) $x lo g x - x + C$
(d) $\frac{( l o g x ) ^{2}}{2} + C$
Hint
Take 1 as the second function.
Answer
(b). With $u = lo g x$ , $v = x$ : $\frac{x ^{2}}{2} lo g x - \int \frac{1}{x} \cdot \frac{x ^{2}}{2} d x = \frac{x ^{2}}{2} lo g x - \frac{x ^{2}}{4} + C$ .
Q2 · 2 marks · Short answerBoard level
Evaluate $\int x sin 3 x d x$ .
Hint
$u = x$ .
Answer
$- \frac{x c o s 3 x}{3} + \int \frac{c o s 3 x}{3} d x = - \frac{x c o s 3 x}{3} + \frac{s i n 3 x}{9} + C$ .
Q3 · 3 marks · Short answerHarder board level
Evaluate $\int x tan^{- 1} x d x$ .
Hint
ILATE: $u = tan^{- 1} x$ . Then write $\frac{x ^{2}}{1 + x ^{2}} = 1 - \frac{1}{1 + x ^{2}}$ .
Answer
$\frac{x ^{2}}{2} tan^{- 1} x - \frac{1}{2} \int \frac{x ^{2}}{1 + x ^{2}} d x = \frac{x ^{2}}{2} tan^{- 1} x - \frac{1}{2} (x - tan^{- 1} x) + C = \frac{x ^{2} + 1}{2} tan^{- 1} x - \frac{x}{2} + C$ .
Q4 · 2 marks · Short answerHarder board level
Evaluate $\int e^{x} (sin x + cos x) d x$ .
Hint
Is the bracket $f + f^{'}$ ?
Answer
With $f = sin x$ , $f^{'} = cos x$ : $e^{x} sin x + C$ .
Q5 · 3 marks · Short answerHarder board level
Evaluate $\int 4 - x^{2} d x$ .
Hint
Use the standard result with $a = 2$ .
Answer
$\frac{x}{2} 4 - x^{2} + 2 sin^{- 1} \frac{x}{2} + C$ .
Q6 · 3 marks · Short answerHOTS / competency
Evaluate $\int \frac{x e ^{x}}{( 1 + x ) ^{2}} d x$ .
Hint
Write $\frac{x}{( 1 + x ) ^{2}} = \frac{( 1 + x ) - 1}{( 1 + x ) ^{2}}$ .
Answer
$\frac{x}{( 1 + x ) ^{2}} = \frac{1}{1 + x} - \frac{1}{( 1 + x ) ^{2}} = f + f^{'}$ with $f = \frac{1}{1 + x}$ . So the integral is $\frac{e ^{x}}{1 + x} + C$ .
Q7 · 4 marks · Case-basedHOTS / competency
Case study: Suppose the concentration of a medicine in the blood after $t$ hours is $C (t) = 10 t e^{- t}$ units.
(i) Find $\int 10 t e^{- t} d t$ . (1 mark)
(ii) Find the total exposure $\int_{0}^{2} C (t) d t$ over the first 2 hours. (1 mark)
(iii) Find when the concentration is greatest. (1 mark)
(iv) Find the greatest concentration. (1 mark)
Hint
For (iii), set $C^{'} (t) = 0$ .
Answer
(i) $- 10 e^{- t} (t + 1) + K$ . (ii) $10 [- e^{- t} (t + 1)]_{0}^{2} = 10 (1 - 3 e^{- 2}) \approx 5.94$ . (iii) $C^{'} (t) = 10 e^{- t} (1 - t) = 0 \Rightarrow t = 1$ h (and $C^{''} (1) < 0$ ). (iv) $C (1) = \frac{10}{e} \approx 3.68$ units.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.State the formula for integration by parts.Show answer
$\int uv d x = u \int v d x - \int (u^{'} \int v d x) d x$ .
2.What does ILATE stand for?Show answer
Inverse trigonometric, Logarithmic, Algebraic, Trigonometric, Exponential.
3.Find $\int e^{x} (tan x + sec^{2} x) d x$ .Show answer
$e^{x} tan x + C$ .
4.Which is the first function in $\int x^{3} lo g x d x$ ?Show answer
$lo g x$ .
5.Find $\int lo g x d x$ .Show answer
$x lo g x - x + C$ .
6.What is $\int (lo g x)^{2} d x$ ?Show answer
$x (lo g x)^{2} - 2 x lo g x + 2 x + C$ .

Integration by Parts: frequently asked questions

How do I decide which function is first in integration by parts?

Use ILATE: inverse trigonometric, logarithmic, algebraic, trigonometric, exponential. The function whose type comes first in this list is the first function u, because it becomes simpler when differentiated.

What if there is only one function, like log x?

Treat it as a product with 1. Take log x as the first function and 1 as the second; the integral of 1 is x, which cancels the 1/x from the derivative of log x.

What is a cyclic integral?

It is an integral such as ∫eˣ sin x dx where applying parts twice brings back the original integral. Call it I, move the I terms to one side and solve for I.

When can I use ∫eˣ[f(x) + f′(x)] dx = eˣf(x) + C?

Whenever the integrand is eˣ multiplied by a function plus its own derivative. Rewrite the bracket to check, for example x/(1 + x)² = 1/(1 + x) − 1/(1 + x)².

How many marks does integration by parts carry in ISC?

It appears regularly as a 4-mark question (for example ∫x² cos x dx in 2024 and ∫(log x)² dx in 2025) and in 1-mark items in Section A. Calculus as a whole carries 35 of the 80 marks in the 2027 paper.