JEE Mains Maths · Indefinite Integration

Integration by Parts and Reverse Differentiation

Integration by parts, the eˣ(f + f′) pattern, and integrands that are already the derivative of a product or a quotient.

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Why this matters

Twelve PYQs, eleven of them multiple choice, and three from 2026. Three integrate by parts, repeatedly or as a reduction formula; four are eˣ(f + f′) in disguise, one of them after putting t = eᵘ; five are the derivative of a product or a quotient, found by guessing its shape and differentiating the guess. Three ideas cover the page.

Concept 1 of 3: Integration by parts

Parts trades one integral for another: differentiate one factor and integrate the other. Choose to differentiate the factor that gets simpler, such as a power of

x

or a log. For a polynomial times

\sin x

e^x

, repeat until the polynomial is gone. For high powers of

\sec x

\csc x

, the original integral comes back, and you solve for it.

Definition

$\int u\,dv=uv-\int v\,du$ . Take $u$ in the order: inverse trig, log, algebraic, trig, exponential.
A polynomial times $\sin x$ , $\cos x$ or $e^x$ : repeat; the signs alternate.
$\int\sec^nx\,dx$ or $\int\csc^nx\,dx$ : take out $\sec^2x$ (or $\csc^2x$ ), use parts, and solve for the integral that returns.
$\int\csc x\,dx=\ln\left|\tan\frac x2\right|$ and $\int\sec x\,dx=\ln|\sec x+\tan x|$ .

Integration by parts

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

Worked example

Find

\int x^2e^x\,dx

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 23 Jan 2025 · Q129Moderate

Example 1 · Indefinite Integration · Integration by Parts and Reverse Differentiation

Let

\int x^{3}\sin x\,dx = g(x)+C

, where

C

is the constant of integration. If

8\left(g\left(\frac{\pi}{2}\right)+g'\left(\frac{\pi}{2}\right)\right)=\alpha\pi^{3}+\beta\pi^{2}+\gamma,\ \alpha,\beta,\gamma\in Z

, then

\alpha+\beta-\gamma

equals:

Signs alternate

Repeated parts alternates the signs:

\int x^2\cos x\,dx=x^2\sin x+2x\cos x-2\sin x

. One wrong sign changes every value computed from the answer.

Drill 2 more on integration by parts

Concept 2 of 3: The pattern eˣ(f + f′)

The derivative of

e^xf(x)

e^x(f+f')

. So when

e^x

multiplies a bracket, look for a function and its derivative inside it. Fractions often hide the pattern: rewrite the top in terms of the bottom, and the two pieces appear. If the variable sits inside a log, put

t=e^u

first to bring out the

e^u

Definition

$\int e^x\left(f(x)+f'(x)\right)dx=e^xf(x)+C$ .
Split a fraction so one piece is the derivative of the other.
A function of $\ln t$ : put $t=e^u$ , $dt=e^u\,du$ .
More generally, $\int e^{kx}\left(kf+f'\right)dx=e^{kx}f+C$ .

The eˣ pattern

\int e^{x}\big(f(x)+f'(x)\big)\,dx=e^{x}f(x)+C

Worked example

Find

\int e^x\left(\frac1x-\frac1{x^2}\right)dx

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 24 Jan 2026 Shift 1 · Q61Moderate

Example 2 · Indefinite Integration · Integration by Parts and Reverse Differentiation

Let

f(t) = \int\left( \frac{1 - \sin\left( \log_{e}t \right)}{1 - \cos\left( \log_{e}t \right)} \right)dt,t > 1

. If

f\left( e^{\pi/2} \right)= -e^{\pi/2}

and

f\left( e^{\pi/4} \right)= \alpha e^{\pi/4}

, then

\alpha

equals

Which piece is f

e^x\left(\frac1x-\frac1{x^2}\right)

f=\frac1x

, because its derivative is the other piece. Taking

f=-\frac1{x^2}

needs

f'=\frac2{x^3}

, which is not there.

Drill 3 more on the pattern eˣ(f + f′)

Concept 3 of 3: Spotting a product or quotient derivative

Some integrands are already a derivative, and the fastest route is to guess the answer and check it. A squared denominator suggests a quotient; a high power of

\sin x

below the line suggests a reciprocal power one lower. Guess the shape, differentiate it, and compare every term. A guess is right only if the derivative matches the integrand exactly.

Definition

$(uv)'=u'v+uv'$ and $\left(\frac uv\right)'=\frac{u'v-uv'}{v^2}$ .
$\frac{\ldots}{\sin^nx\cos^mx}$ : try $\frac{1}{\sin^{n-1}x\cos^{m-1}x}$ or $\frac{\tan x}{\sin^nx}$ .
$e^{g(x)}$ times a bracket: try $e^{g}h$ , whose derivative is $e^{g}(g'h+h')$ .
$\left(\frac xa\right)^x$ has derivative $\left(\frac xa\right)^x\left(1+\ln\frac xa\right)$ .

Quotient in reverse

\int\frac{u'v-uv'}{v^2}\,dx=\frac{u}{v}+C

Worked example

Find

\int\frac{x\cos x-\sin x}{x^2}\,dx

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 28 Jan 2026 Shift 1 · Q61Moderate

Example 3 · Indefinite Integration · Integration by Parts and Reverse Differentiation

\int_{}^{}\ \left( \frac{1 - 5\cos^{2}x}{\sin^{5}x\cos^{2}x} \right)dx=f(x) +C

where

C

is the constant of integration, then

f\left( \frac{\pi}{6} \right)- f\left( \frac{\pi}{4} \right)

is equal to

Differentiate the guess

A guess that is close is not an antiderivative. Differentiate it and compare term by term:

\ln x

and

x-1

agree at

x=1

and nowhere else, so one cannot stand in for the other.

Drill 4 more on spotting a product or quotient derivative

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)

Integration by parts
Integration by parts
$\int u\,dv=uv-\int v\,du$
The pattern eˣ(f + f′)
The eˣ pattern
$\int e^{x}\big(f(x)+f'(x)\big)\,dx=e^{x}f(x)+C$
Spotting a product or quotient derivative
Quotient in reverse
$\int\frac{u'v-uv'}{v^2}\,dx=\frac{u}{v}+C$

Watch out for (3)

Signs alternate
→ Integration by parts
Which piece is f
→ The pattern eˣ(f + f′)
Differentiate the guess
→ Spotting a product or quotient derivative

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.

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