How does the e^x[f(x) + f prime(x)] shortcut work?

Whenever the integrand has e^x multiplied by a bracket equal to some function plus its derivative, the integral equals e^x times f(x) plus C. This follows from integration by parts. The skill is spotting the pattern - the bracket often needs algebraic rearrangement first.

How do I integrate ln x dx?

Take u = ln x and v = 1. Then du = dx/x and the integral of v dx equals x. So the integral of ln x dx equals x ln x minus x plus C. The same trick works for arcsin x and arctan x.

Are all three formulae for sqrt(x^2 plus or minus a^2) derivable by parts?

Yes. All three follow from the same cyclic-parts calculation, with signs adjusted for the plus or minus a^2 term. They are standard results memorised for JEE and appear in problems on areas, arc length, and reduction formulae.

Integration by Parts

Q: What is the integration by parts formula?

The integral of u dv equals uv minus the integral of v du. It comes from integrating the product rule for derivatives.

Q: What is cyclic integration by parts?

When applying integration by parts twice, the original integral I reappears with a coefficient like -1 on the right-hand side. Move it to the left, giving 2I equals a closed-form expression, and divide by 2. Also called solving for the integral algebraically.

Q: Can I use integration by parts more than once?

Yes. For integrands like x^2 e^x or x^2 sin x, apply by-parts twice - once to reduce the power of x from 2 to 1, then again to eliminate it. For x^n e^x you would apply it n times.

Q: Why does ILATE fail sometimes?

ILATE is a guideline, not a theorem. In special cases like the integral of e^x sin x, either choice of u leads to a cyclic integration. The rule prioritises the most common scenarios; when the standard choice produces a harder integral, swap u and v.

MathsIntegralsFor JEE aspirants

Integration by parts is the reverse of the product rule for differentiation. For two functions $u (x)$ and $v (x)$ , it states $\int u v d x = u \int v d x - \int [\frac{d u}{d x} \int v d x] d x$ . The core skill is choosing which function to treat as $u$ (the "first function"). The standard guide is the ILATE rule: prefer Inverse trig, then Logarithmic, then Algebraic, then Trigonometric, then Exponential. This concept covers ILATE, cyclic integration (where the target integral reappears on the right side and gets solved algebraically), and the powerful shortcut $\int e^{x} [f (x) + f^{'} (x)] d x = e^{x} f (x) + C$ .

Key Formulas - Quick Reference

$\int u v d x = u \int v d x - \int [\frac{d u}{d x} \int v d x] d x$
ILATE order (choose $u$ as the leftmost applicable): I (Inverse trig), L (Log), A (Algebraic), T (Trig), E (Exponential)
$\int e^{x} [f (x) + f^{'} (x)] d x = e^{x} f (x) + C$
$\int x^{2} + a^{2} d x = \frac{x}{2} x^{2} + a^{2} + \frac{a ^{2}}{2} ln x + x^{2} + a^{2} + C$
$\int x^{2} - a^{2} d x = \frac{x}{2} x^{2} - a^{2} - \frac{a ^{2}}{2} ln x + x^{2} - a^{2} + C$
$\int a^{2} - x^{2} d x = \frac{x}{2} a^{2} - x^{2} + \frac{a ^{2}}{2} sin^{- 1} \frac{x}{a} + C$
$\int sec^{3} x d x = \frac{1}{2} sec x tan x + \frac{1}{2} ln ∣ sec x + tan x ∣ + C$
$\int e^{a x} sin (b x) d x = \frac{e ^{a x} [ a sin ( b x ) - b cos ( b x ) ]}{a ^{2} + b ^{2}} + C$
$\int e^{a x} cos (b x) d x = \frac{e ^{a x} [ a cos ( b x ) + b sin ( b x ) ]}{a ^{2} + b ^{2}} + C$

1. The Formula and Its Derivation

If $u (x)$ and $v (x)$ are differentiable, the product rule states $\frac{d}{d x} [u \cdot w] = \frac{d u}{d x} \cdot w + u \cdot \frac{d w}{d x}$ , where $w = \int v d x$ . Integrating both sides:

$u \cdot w = \int \frac{d u}{d x} \cdot w d x + \int u \cdot \frac{d w}{d x} d x = \int \frac{d u}{d x} (\int v d x) d x + \int u \cdot v d x .$

Rearranging gives the integration by parts formula:

$\int u v d x = u (\int v d x) - \int [\frac{d u}{d x} (\int v d x)] d x .$

In shorthand (with $d v = v d x$ ): $\int u d v = uv - \int v d u$ .

2. Choosing $u$ : The ILATE Rule

The success of integration by parts hinges on choosing $u$ so that $\frac{d u}{d x}$ is simpler than $u$ itself, and $\int v d x$ is easy. The ILATE rule gives a reliable ordering: choose $u$ as the function appearing leftmost in the following list, and $v$ as everything else.

ILATE:

I - Inverse trigonometric ( $sin^{- 1} x$ , $tan^{- 1} x$ , etc.)
L - Logarithmic ( $ln x$ , $lo g x$ )
A - Algebraic (polynomials, rational functions like $x, x^{2}, 1/ x$ )
T - Trigonometric ( $sin x, cos x, tan x$ )
E - Exponential ( $e^{x}, a^{x}$ )

Special cases:

If one of the two functions is not directly integrable (e.g. $ln x, sin^{- 1} x, tan^{- 1} x$ ), always pick it as $u$ - since it needs to be differentiated, not integrated.
If there is only one function (e.g. $\int tan^{- 1} x d x$ ), take $u = tan^{- 1} x$ and $v = 1$ . So $\int v d x = x$ .
For $\int x sin x d x$ : by ILATE, algebraic beats trig, so $u = x, v = sin x$ .
For $\int x tan^{- 1} x d x$ : inverse beats algebraic, so $u = tan^{- 1} x, v = x$ .

3. Cyclic Integration by Parts

Sometimes applying integration by parts once (or twice) makes the original integral reappear on the right-hand side. Instead of an infinite loop, treat the appearance algebraically - move it to the left side and divide.

Solved Example 1

Evaluate

\int x^{2} + a^{2} d x .

Solution:

Take $u = x^{2} + a^{2}$ and $v = 1$ . Then $\frac{d u}{d x} = \frac{x}{x ^{2} + a ^{2}}$ and $\int v d x = x$ :

$I = x x^{2} + a^{2} - \int \frac{x ^{2}}{x ^{2} + a ^{2}} d x .$

Rewrite $x^{2} = (x^{2} + a^{2}) - a^{2}$ :

$I = x x^{2} + a^{2} - \int \frac{x ^{2} + a ^{2}}{x ^{2} + a ^{2}} d x + \int \frac{a ^{2}}{x ^{2} + a ^{2}} d x .$

$I = x x^{2} + a^{2} - \int x^{2} + a^{2} d x + a^{2} ln x + x^{2} + a^{2} .$

The middle term is $- I$ . Moving it to the left:

$2 I = x x^{2} + a^{2} + a^{2} ln x + x^{2} + a^{2} + const .$

$I = \frac{x}{2} x^{2} + a^{2} + \frac{a ^{2}}{2} ln x + x^{2} + a^{2} + C .$

Solved Example 2

Evaluate

\int sec^{3} θ d θ .

Solution:

Write $sec^{3} θ = sec θ \cdot sec^{2} θ$ . Take $u = sec θ, v = sec^{2} θ$ . Then $d u = sec θ tan θ d θ$ and $\int v d θ = tan θ$ :

$I = sec θ tan θ - \int tan θ (sec θ tan θ) d θ = sec θ tan θ - \int sec θ tan^{2} θ d θ .$

Use $tan^{2} θ = sec^{2} θ - 1$ :

$I = sec θ tan θ - \int (sec^{3} θ - sec θ) d θ = sec θ tan θ - I + ln ∣ sec θ + tan θ ∣.$

Move $- I$ to the left:

$2 I = sec θ tan θ + ln ∣ sec θ + tan θ ∣ \Rightarrow I = \frac{1}{2} sec θ tan θ + \frac{1}{2} ln ∣ sec θ + tan θ ∣ + C .$

4. Direct Applications

Solved Example 3

Evaluate

\int x^{3} ln x d x .

Solution:

By ILATE, $ln x$ (Logarithmic) comes before $x^{3}$ (Algebraic), so $u = ln x, v = x^{3}$ .

$d u = \frac{1}{x} d x, \int v d x = \frac{x ^{4}}{4} .$

$I = ln x \cdot \frac{x ^{4}}{4} - \int \frac{x ^{4}}{4} \cdot \frac{1}{x} d x = \frac{x ^{4}}{4} ln x - \frac{1}{4} \int x^{3} d x = \frac{x ^{4}}{4} ln x - \frac{x ^{4}}{16} + C .$

5. The $e^{x} [f (x) + f^{'} (x)]$ Shortcut

An extremely useful identity: if the integrand has the shape $e^{x}$ times a bracket that equals a function plus its derivative, the answer is one line.

$\int e^{x} [f (x) + f^{'} (x)] d x = e^{x} f (x) + C .$

Proof (by parts): $\int e^{x} f (x) d x = e^{x} f (x) - \int e^{x} f^{'} (x) d x$ , so $\int e^{x} [f (x) + f^{'} (x)] d x = e^{x} f (x)$ .

Solved Example 4

Evaluate

\int e^{x} (\frac{1}{x} - \frac{1}{x ^{2}}) d x .

Solution:

Let $f (x) = \frac{1}{x}$ . Then $f^{'} (x) = - \frac{1}{x ^{2}}$ . The bracket $\frac{1}{x} - \frac{1}{x ^{2}} = f (x) + f^{'} (x)$ , so by the shortcut:

$I = e^{x} \cdot \frac{1}{x} + C = \frac{e ^{x}}{x} + C .$

Solved Example 5

Evaluate

\int e^{x} (\frac{1 - x}{1 + x ^{2}})^{2} d x .

Solution:

Expand $(1 - x)^{2} = 1 - 2 x + x^{2}$ , so

$\frac{( 1 - x ) ^{2}}{( 1 + x ^{2} ) ^{2}} = \frac{1 + x ^{2} - 2 x}{( 1 + x ^{2} ) ^{2}} = \frac{1}{1 + x ^{2}} - \frac{2 x}{( 1 + x ^{2} ) ^{2}} .$

Let $f (x) = \frac{1}{1 + x ^{2}}$ . Then $f^{'} (x) = - \frac{2 x}{( 1 + x ^{2} ) ^{2}}$ . So the bracket is $f (x) + f^{'} (x)$ , and

$I = e^{x} \cdot \frac{1}{1 + x ^{2}} + C = \frac{e ^{x}}{1 + x ^{2}} + C .$

6. Exponential Times Trigonometric: $\int e^{a x} sin (b x) d x$ and $\int e^{a x} cos (b x) d x$

These integrals are the most frequently tested cyclic-parts examples in JEE Mains. Applying integration by parts twice makes the original integral reappear on the right, and it can be solved algebraically. The general result:

$\int e^{a x} sin (b x) d x = \frac{e ^{a x} [ a sin ( b x ) - b cos ( b x ) ]}{a ^{2} + b ^{2}} + C .$

$\int e^{a x} cos (b x) d x = \frac{e ^{a x} [ a cos ( b x ) + b sin ( b x ) ]}{a ^{2} + b ^{2}} + C .$

Both formulas can be derived by cyclic integration by parts, as shown below.

Solved Example 6

Derive

\int e^{a x} sin (b x) d x .

Solution:

Let $I = \int e^{a x} sin (b x) d x$ . Take $u = sin (b x), v = e^{a x}$ . Then $d u = b cos (b x) d x$ and $\int v d x = \frac{e ^{a x}}{a}$ :

$I = \frac{e ^{a x} sin ( b x )}{a} - \frac{b}{a} \int e^{a x} cos (b x) d x .$

Apply parts again to the remaining integral with $u = cos (b x), v = e^{a x}$ :

$\int e^{a x} cos (b x) d x = \frac{e ^{a x} cos ( b x )}{a} + \frac{b}{a} \int e^{a x} sin (b x) d x = \frac{e ^{a x} cos ( b x )}{a} + \frac{b}{a} I .$

Substituting back:

$I = \frac{e ^{a x} sin ( b x )}{a} - \frac{b}{a} [\frac{e ^{a x} cos ( b x )}{a} + \frac{b}{a} I] = \frac{e ^{a x} sin ( b x )}{a} - \frac{b e ^{a x} cos ( b x )}{a ^{2}} - \frac{b ^{2}}{a ^{2}} I .$

Collect $I$ on the left:

$I + \frac{b ^{2}}{a ^{2}} I = \frac{a e ^{a x} sin ( b x ) - b e ^{a x} cos ( b x )}{a ^{2}} \Rightarrow \frac{a ^{2} + b ^{2}}{a ^{2}} I = \frac{e ^{a x} [ a sin ( b x ) - b cos ( b x ) ]}{a ^{2}} .$

$I = \frac{e ^{a x} [ a sin ( b x ) - b cos ( b x ) ]}{a ^{2} + b ^{2}} + C .$

The companion formula for $\int e^{a x} cos (b x) d x$ follows identically, with the numerator becoming $a cos (b x) + b sin (b x)$ .

Common Mistakes to Avoid

Watch out

Choosing $u$ and $v$ against ILATE. Picking the "wrong" function as $u$ often produces an integral harder than the original. If you're stuck partway through, swap $u$ and $v$ and restart.
Forgetting the $- \int v d u$ term. The formula is $\int u d v = uv - \int v d u$ . The minus sign is essential; dropping it produces answers off by twice the correction term.
In cyclic integration, not recognising the original integral has returned. If after applying by-parts once or twice the integral $I$ reappears on the right (typically with a coefficient of $\pm 1$ ), stop and solve algebraically: $2 I = \dots \Rightarrow I = \frac{1}{2} (\dots)$ .
Missing the $e^{x} [f + f^{'}]$ pattern. Any integrand of the form $e^{x} \cdot [something]$ - especially with $\frac{1}{x} - \frac{1}{x ^{2}}$ or $tan x + sec^{2} x$ style brackets - is a candidate. Always test if the bracket equals $f + f^{'}$ for some obvious $f$ .
For $\int ln x d x$ , forgetting $v = 1$ . Even a single function can be integrated by parts by taking the second function as 1: $\int ln x d x = x ln x - \int x \cdot \frac{1}{x} d x = x ln x - x + C$ .

Frequently Asked Questions

Q1. What is the integration by parts formula?

$\int u d v = uv - \int v d u$ , or equivalently $\int u v d x = u \int v d x - \int [\frac{d u}{d x} \int v d x] d x$ . It comes from integrating the product rule for derivatives.

Q2. What is the ILATE rule and why does it work?

ILATE (Inverse trig, Logarithmic, Algebraic, Trigonometric, Exponential) gives the priority order for choosing $u$ - pick the function class appearing leftmost. It works because differentiating an inverse-trig or log function simplifies it (removes the transcendental piece), while integrating trig or exponential leaves the class unchanged - so the resulting $\int v d u$ is easier.

Q3. How does the $e^{x} [f (x) + f^{'} (x)]$ shortcut work?

Whenever the integrand has $e^{x}$ multiplied by a bracket equal to some function plus its derivative, $\int e^{x} [f (x) + f^{'} (x)] d x = e^{x} f (x) + C$ . This falls out immediately from integration by parts. The skill is spotting the pattern - the bracket often needs algebraic rearrangement first.

Q4. What is cyclic integration by parts?

When applying integration by parts twice (usually to $\int e^{x} sin x$ or $\int sec^{3} x$ or $\int x^{2} + a^{2}$ ), the original integral $I$ reappears with a coefficient like $- 1$ on the right-hand side. Move it to the left, giving $2 I = (no integral)$ , and divide by 2. Also called "solving for the integral algebraically".

Q5. Can I use integration by parts more than once?

Yes. For integrands like $x^{2} e^{x}$ or $x^{2} sin x$ , apply by-parts twice (once to reduce the power of $x$ from 2 to 1, then again to eliminate it). For $x^{n} e^{x}$ you would apply it $n$ times, but a shortcut is $\int x^{n} e^{x} d x = e^{x} (x^{n} - n x^{n - 1} + n (n - 1) x^{n - 2} - \dots)$ .

Q6. How do I integrate $\int ln x d x$ ?

Take $u = ln x$ and $v = 1$ . Then $d u = d x / x$ and $\int v d x = x$ . So $\int ln x d x = x ln x - \int x \cdot \frac{1}{x} d x = x ln x - x + C$ . The same trick works for $\int sin^{- 1} x d x$ and $\int tan^{- 1} x d x$ .

Q7. Why does ILATE fail sometimes?

ILATE is a guideline, not a theorem. In special cases (like $\int e^{x} sin x d x$ , where I and E don't apply and both L and T are absent), either choice of $u$ leads to a cyclic integration. The rule prioritises the most common scenarios; when the standard choice produces a harder integral, swap.

Q8. Are all three formulae $\int x^{2} \pm a^{2} d x$ derivable by parts?

Yes. All three follow from the same cyclic-parts calculation shown in Solved Example 1, with signs adjusted for the $\pm a^{2}$ . They are standard results memorised for JEE and appear directly in problems on areas, arc length, and reduction formulae.

Previous year questions on Integration by Parts

9 questions from past papers, each with a step-by-step solution.

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Practice questions →Take a mock test

1. The Formula and Its Derivation

2. Choosing u: The ILATE Rule

3. Cyclic Integration by Parts

4. Direct Applications

5. The ex[f(x)+f′(x)] Shortcut

6. Exponential Times Trigonometric: ∫eaxsin(bx)dx and ∫eaxcos(bx)dx

Common Mistakes to Avoid

Frequently Asked Questions

Q1. What is the integration by parts formula?

Q2. What is the ILATE rule and why does it work?

Q3. How does the ex[f(x)+f′(x)] shortcut work?

Q4. What is cyclic integration by parts?

Q5. Can I use integration by parts more than once?

Q6. How do I integrate ∫lnxdx?

Q7. Why does ILATE fail sometimes?

Q8. Are all three formulae ∫x2±a2​dx derivable by parts?

Previous year questions on Integration by Parts

Ready to master Integrals?

2. Choosing $u$ : The ILATE Rule

5. The $e^{x} [f (x) + f^{'} (x)]$ Shortcut

6. Exponential Times Trigonometric: $\int e^{a x} sin (b x) d x$ and $\int e^{a x} cos (b x) d x$

Q3. How does the $e^{x} [f (x) + f^{'} (x)]$ shortcut work?

Q6. How do I integrate $\int ln x d x$ ?

Q8. Are all three formulae $\int x^{2} \pm a^{2} d x$ derivable by parts?