The Exponential "Magic" Shortcut

🎯 The Strategy

This is the most frequently tested shortcut in JEE Main for indefinite integration. Whenever you see $e^x$ combined with a function and its derivative, you can apply these powerful identities directly.

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📐 The Core Identities

Identity 1: The Classic $e^x$ Formula

$\boxed{\int e^x [f(x) + f'(x)] \, dx = e^x f(x) + C}$

This works because: $\frac{d}{dx}[e^x f(x)] = e^x f(x) + e^x f'(x) = e^x[f(x) + f'(x)]$

Identity 2: The $xf(x)$ Extension

$\boxed{\int [f(x) + x f'(x)] \, dx = x f(x) + C}$

This is essentially the product rule in reverse: $\frac{d}{dx}[xf(x)] = f(x) + xf'(x)$

Identity 3: The General Exponential Form

$\boxed{\int e^{g(x)} [f(x) \cdot g'(x) + f'(x)] \, dx = e^{g(x)} f(x) + C}$

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🔍 Pattern Recognition

Look for these telltale signs:

• $e^x$ multiplied by a sum of two terms
• One term looks like a function, the other looks like its derivative
• The terms share similar structure (same denominator, related powers)

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📝 JEE Main Previous Year Examples

Example 1: JEE Main 2019

Question: If $\int e^x\left(\frac{1 + \sin x}{1 + \cos x}\right)dx = e^x \cdot f(x) + C$, find $f(x)$.

Solution:

First, simplify the integrand using half-angle formulas:

$\frac{1 + \sin x}{1 + \cos x} = \frac{1 + 2\sin(x/2)\cos(x/2)}{2\cos^2(x/2)}$

$= \frac{1}{2\cos^2(x/2)} + \frac{2\sin(x/2)\cos(x/2)}{2\cos^2(x/2)}$

$= \frac{1}{2}\sec^2(x/2) + \tan(x/2)$

Now the integral becomes: $\int e^x\left[\tan(x/2) + \frac{1}{2}\sec^2(x/2)\right]dx$

Recognize the pattern!

• Let $f(x) = \tan(x/2)$
• Then $f'(x) = \frac{1}{2}\sec^2(x/2)$

This matches: $\int e^x[f(x) + f'(x)]dx = e^x f(x) + C$

Answer: $f(x) = \tan(x/2)$

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Example 2: JEE Main 2021

Question: If $\int (e^{2x} + 2e^x - e^{-x} - 1)e^{(e^x + e^{-x})}dx = g(x)e^{(e^x + e^{-x})} + C$, where $C$ is a constant of integration, then $g(0)$ is equal to:

Solution:

Let $u = e^x + e^{-x}$, so $\frac{du}{dx} = e^x - e^{-x}$

The integrand can be rewritten: $e^{2x} + 2e^x - e^{-x} - 1 = (e^{2x} - 1) + (2e^x - e^{-x})$

$= (e^x - e^{-x})(e^x + 1) + e^x$

Hmm, let's try another approach. Factor out strategically:

$(e^{2x} + 2e^x - e^{-x} - 1) = (e^x - e^{-x})(e^x + 1) + e^x$

Using Identity 3 with $g(x) = e^x + e^{-x}$: $g'(x) = e^x - e^{-x}$

We need: $f(x) \cdot g'(x) + f'(x) = e^{2x} + 2e^x - e^{-x} - 1$

If $f(x) = e^x$, then:

• $f'(x) = e^x$
• $f(x) \cdot g'(x) = e^x(e^x - e^{-x}) = e^{2x} - 1$

So: $f(x)g'(x) + f'(x) = e^{2x} - 1 + e^x$

This doesn't match exactly. Let's try $f(x) = e^x + 1$:

• $f'(x) = e^x$
• $f(x) \cdot g'(x) = (e^x + 1)(e^x - e^{-x}) = e^{2x} - 1 + e^x - e^{-x}$

So: $f(x)g'(x) + f'(x) = e^{2x} + 2e^x - e^{-x} - 1$ ✓

Therefore: $g(x) = e^x + 1$

Answer: $g(0) = e^0 + 1 = 2$

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Example 3: JEE Main 2025

Question: The integral $\int \left(1 + x - \frac{1}{x}\right)e^{x + \frac{1}{x}}dx$ is equal to:

Options: (A) $(x + 1)e^{x + \frac{1}{x}} + C$ (B) $-xe^{x + \frac{1}{x}} + C$ (C) $(x - 1)e^{x + \frac{1}{x}} + C$ (D) $xe^{x + \frac{1}{x}} + C$

Solution:

Let $g(x) = x + \frac{1}{x}$

Then $g'(x) = 1 - \frac{1}{x^2}$

The integrand is: $\left(1 + x - \frac{1}{x}\right)e^{x + \frac{1}{x}}$

We need to express this as $[f(x) \cdot g'(x) + f'(x)]e^{g(x)}$

$1 + x - \frac{1}{x} = x\left(1 - \frac{1}{x^2}\right) + 1 = x \cdot g'(x) + 1$

If $f(x) = x$, then $f'(x) = 1$

So: $f(x) \cdot g'(x) + f'(x) = x\left(1 - \frac{1}{x^2}\right) + 1 = x - \frac{1}{x} + 1$ ✓

By Identity 3: $\int \left(1 + x - \frac{1}{x}\right)e^{x + \frac{1}{x}}dx = x \cdot e^{x + \frac{1}{x}} + C$

Answer: (D)

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Example 4: JEE Main 2018

Question: If $\int x^5 e^{-4x^3}dx = \frac{1}{48}e^{-4x^3}f(x) + C$, where $C$ is a constant of integration, then $f(x)$ is equal to:

Options: (A) $-4x^3 - 1$ (B) $4x^3 + 1$ (C) $-2x^3 - 1$ (D) $-2x^3 + 1$

Solution:

Let $g(x) = -4x^3$, so $g'(x) = -12x^2$

Rewrite the integrand: $x^5 e^{-4x^3} = x^5 \cdot e^{g(x)}$

We need $x^5 = f(x) \cdot g'(x) + f'(x) = -12x^2 f(x) + f'(x)$

Try $f(x) = ax^3 + b$ (polynomial guess based on structure):

• $f'(x) = 3ax^2$
• $f(x) \cdot g'(x) = -12x^2(ax^3 + b) = -12ax^5 - 12bx^2$

So: $-12ax^5 - 12bx^2 + 3ax^2 = x^5$

Comparing coefficients:

• $x^5$: $-12a = 1 \Rightarrow a = -\frac{1}{12}$
• $x^2$: $-12b + 3a = 0 \Rightarrow b = \frac{a}{4} = -\frac{1}{48}$

So $f(x) = -\frac{1}{12}x^3 - \frac{1}{48}$

The integral is: $\frac{1}{48}e^{-4x^3} \cdot (-4x^3 - 1) + C$

Answer: (A) $f(x) = -4x^3 - 1$

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Example 5: JEE Main 2019

Question: $\int e^{\sec x}(\sec x \tan x f(x) + \sec x \tan x + \sec^2 x)dx = e^{\sec x}f(x) + C$. Find $f(x)$.

Solution:

Let $g(x) = \sec x$, so $g'(x) = \sec x \tan x$

The integrand is: $e^{\sec x}[\sec x \tan x \cdot f(x) + \sec x \tan x + \sec^2 x]$ $= e^{g(x)}[g'(x) \cdot f(x) + \sec x \tan x + \sec^2 x]$

For this to match Identity 3, we need: $f'(x) = \sec x \tan x + \sec^2 x$

Integrating: $f(x) = \sec x + \tan x$

Verification:

• $f(x) = \sec x + \tan x$
• $f'(x) = \sec x \tan x + \sec^2 x$ ✓

Answer: $f(x) = \sec x + \tan x$

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🔑 Quick Recognition Guide

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💡 How to Spot the Pattern

1. Look for $e^{(\text{something})}$ in the integrand
2. Identify what's being multiplied with the exponential
3. Check if the sum has a function and its derivative
4. Verify by differentiation if unsure

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⏱️ Time Saved

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📌 Practice Problems

1. $\int e^x(\sin x + \cos x)dx$ Hint: $\frac{d}{dx}(\sin x) = \cos x$

2. $\int e^x\left(\frac{x-1}{x^2}\right)dx$ Hint: Separate and identify $f(x)$

3. $\int (1 + x\ln x)dx$ Hint: Use Identity 2

4. $\int e^{\tan x}(\sec^2 x + \sec^2 x \tan x)dx$ Hint: $g(x) = \tan x$