Differentiation in Reverse: The "Options" Trick for JEE Integration

🎯 The Strategic Edge

The Core Insight: In the MCQ format of JEE Mains, you have a powerful tool at your disposal: the answer choices themselves. When faced with a complex or time-consuming integration problem, you can often find the correct answer by differentiating the options instead of integrating the given function.

$\text{If } \int f(x)\,dx = F(x) + C, \text{ then } \boxed{\frac{d}{dx}[F(x)] = f(x)}$

This reverse approach transforms a challenging integration into a simpler, often mechanical, differentiation problem.

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📐 The Method: Step-by-Step

When to Use This Strategy:

✅ The question asks "find the integral of..." ✅ You have 4-5 distinct function options. ✅ Direct integration looks messy or time-consuming. ✅ You're short on time or unsure of the integration technique.

The 3-Step Process:

1. Scan the Options: Quickly look at the structure of each option (logarithms, inverse trig, rational functions, etc.).
2. Differentiate Each Option: Compute $\frac{d}{dx}[\text{Option}]$.
3. Match with the Integrand: Compare the derivative to the original $f(x)$ given in the problem.

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⚡ Pro Techniques for Maximum Speed

Technique 1: Smart Elimination via Coefficient Check

You often don't need to differentiate completely. Check:

• The leading coefficient after differentiation.
• The degree/power of the resulting expression.

Example: If differentiating an option yields $3x^2$ but the integrand has $6x^2$, you know the correct option might have a constant factor of 2.

Technique 2: Strategic Substitution (The "Spot Check")

Pick a simple value of $x$ (like $x = 0, 1, \pi/4$) and:

1. Compute $f(x)$ at that point.
2. Differentiate each option at that same $x$ value.
3. Eliminate options where $F'(x) \neq f(x)$.

This is especially fast for trigonometric integrals.

Technique 3: Structure Recognition

Match the derivative's structure to the integrand's structure:

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📝 Applied Mastery: JEE Problems Decoded

Problem Type 1: The Standard Verification

Question (JEE Main 2019): If $\int \frac{(x^2 + 1)e^x}{(x+1)^2}dx = f(x)e^x + C$, find $\frac{d^3f}{dx^3}$ at $x = 1$.

Options: (A) $-\frac{3}{4}$ (B) $\frac{3}{4}$ (C) $\frac{1}{2}$ (D) $-\frac{1}{2}$

Solution using Options Trick: We need to find $f(x)$ first. The problem says the integral equals $f(x)e^x + C$. So if we differentiate that, we should get the integrand back: $\frac{d}{dx}[f(x)e^x] = e^x[f(x) + f'(x)] = \frac{(x^2+1)e^x}{(x+1)^2}$ Thus, $f(x) + f'(x) = \frac{x^2+1}{(x+1)^2}$.

Instead of solving this differential equation, let's test intelligent guesses for $f(x)$ based on the options for $f'''(1)$.

We can try a simple rational function: $f(x) = \frac{Ax+B}{x+1}$. Differentiate: $f'(x) = \frac{A(x+1) - (Ax+B)}{(x+1)^2} = \frac{A - B}{(x+1)^2}$.

Then: $f(x) + f'(x) = \frac{Ax+B}{x+1} + \frac{A-B}{(x+1)^2} = \frac{(Ax+B)(x+1) + (A-B)}{(x+1)^2} = \frac{Ax^2 + (A+B)x + B + A - B}{(x+1)^2} = \frac{Ax^2 + (A+B)x + A}{(x+1)^2}$ We need this to equal $\frac{x^2+1}{(x+1)^2}$. So:

• Coefficient of $x^2$: $A = 1$
• Coefficient of $x$: $A+B = 0 \Rightarrow B = -1$
• Constant term: $A = 1$ ✓

Thus $f(x) = \frac{x-1}{x+1}$. Now compute derivatives: $f'(x) = \frac{2}{(x+1)^2},\quad f''(x) = \frac{-4}{(x+1)^3},\quad f'''(x) = \frac{12}{(x+1)^4}$ At $x=1$: $f'''(1) = \frac{12}{16} = \frac{3}{4}$.

Answer: (B) $\frac{3}{4}$.

Time saved: Instead of integrating a tricky rational-exponential function, we solved a simple differential equation by intelligent guessing.

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Problem Type 2: Trigonometric Integral with Inverse Trig Answer

Question (JEE Main 2020): If $\int \frac{\cos x - \sin x}{\sqrt{8 - \sin 2x}}dx = a\sin^{-1}\left(\frac{\sin x + \cos x}{b}\right) + c$, find $(a, b)$.

Options: (A) $(1, 3)$ (B) $(-1, 3)$ (C) $(1, -3)$ (D) $(-1, -3)$

Solution using Differentiation Check: Let's test option (A): $F(x) = \sin^{-1}\left(\frac{\sin x + \cos x}{3}\right)$.

Differentiate using chain rule: $\frac{dF}{dx} = \frac{1}{\sqrt{1 - \left(\frac{\sin x+\cos x}{3}\right)^2}} \cdot \frac{d}{dx}\left(\frac{\sin x+\cos x}{3}\right)$ $= \frac{1}{\sqrt{1 - \frac{(\sin x+\cos x)^2}{9}}} \cdot \frac{\cos x - \sin x}{3}$ Simplify the square root: $\sqrt{1 - \frac{(\sin x+\cos x)^2}{9}} = \sqrt{\frac{9 - (1 + \sin 2x)}{9}} = \sqrt{\frac{8 - \sin 2x}{9}} = \frac{\sqrt{8 - \sin 2x}}{3}$ Therefore: $\frac{dF}{dx} = \frac{1}{(\sqrt{8 - \sin 2x})/3} \cdot \frac{\cos x - \sin x}{3} = \frac{\cos x - \sin x}{\sqrt{8 - \sin 2x}}$ This matches the integrand exactly! So option (A) is correct.

Answer: (A) $(1, 3)$.

Time saved: No need for trigonometric substitution or clever algebraic manipulation of the integrand.

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Problem Type 3: Rational Function Integration

Question (JEE Main 2018): The integral $\int \frac{2x^3 - 1}{x^4 + x}dx$ equals:

Options:
(A) $\frac{1}{2}\ln\left|\frac{x^3 + 1}{x^2}\right| + C$
(B) $\ln\left|\frac{x^3 + 1}{x}\right| + C$
(C) $\frac{1}{2}\ln\left|\frac{x^3 + 1}{x}\right| + C$
(D) $\ln\left|\frac{x^3 + 1}{x^2}\right| + C$

Solution using Spot Check & Differentiation: Let's use $x = 1$ as a test point.

1. Integrand at $x=1$: $\frac{2(1)^3 - 1}{1^4 + 1} = \frac{1}{2}$.
2. Derivative of each option at $x=1$:

For option (A): $F(x) = \frac{1}{2}\ln\left|\frac{x^3+1}{x^2}\right|$ $F'(x) = \frac{1}{2} \cdot \frac{x^2}{x^3+1} \cdot \frac{3x^2 \cdot x^2 - (x^3+1)\cdot 2x}{x^4} = \frac{1}{2} \cdot \frac{3x^4 - 2x(x^3+1)}{x^2(x^3+1)}$ At $x=1$: $F'(1) = \frac{1}{2} \cdot \frac{3 - 2(1+1)}{1 \cdot (1+1)} = \frac{1}{2} \cdot \frac{3-4}{2} = \frac{1}{2} \cdot \frac{-1}{2} = -\frac{1}{4} \neq \frac{1}{2}$ ✗

For option (C): $F(x) = \frac{1}{2}\ln\left|\frac{x^3+1}{x}\right|$ $F'(x) = \frac{1}{2} \cdot \frac{x}{x^3+1} \cdot \frac{3x^2 \cdot x - (x^3+1)\cdot 1}{x^2} = \frac{1}{2} \cdot \frac{3x^3 - (x^3+1)}{x(x^3+1)}$ At $x=1$: $F'(1) = \frac{1}{2} \cdot \frac{3 - (1+1)}{1\cdot(1+1)} = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4} \neq \frac{1}{2}$ ✗

Wait, we made an arithmetic error. Let's recompute carefully for option (C): $F(x) = \frac{1}{2}\ln\left|\frac{x^3+1}{x}\right| = \frac{1}{2}[\ln|x^3+1| - \ln|x|]$ $F'(x) = \frac{1}{2}\left[\frac{3x^2}{x^3+1} - \frac{1}{x}\right]$ At $x=1$: $F'(1) = \frac{1}{2}\left[\frac{3}{2} - 1\right] = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}$. Still not $\frac{1}{2}$.

Let's check option (B): $F(x) = \ln\left|\frac{x^3+1}{x}\right| = \ln|x^3+1| - \ln|x|$ $F'(x) = \frac{3x^2}{x^3+1} - \frac{1}{x}$ At $x=1$: $F'(1) = \frac{3}{2} - 1 = \frac{1}{2}$ ✓

Therefore, option (B) is correct.

Answer: (B) $\ln\left|\frac{x^3 + 1}{x}\right| + C$.

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🔑 Decision Framework: When to Integrate vs. When to Differentiate

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⏱️ Time Analysis & Strategic Advantage

Best Case Scenario: The correct option is the first one you differentiate, or a spot check eliminates 3 options immediately.

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📌 Practice for Speed (Try in 2 minutes each)

1. JEE Main 2021: $\int \frac{x^3 - x}{x^4 + 9}dx = \frac{1}{2}\ln(x^4+9) - \frac{1}{6}\tan^{-1}\left(\frac{x^2}{3}\right) + C$. Verify by differentiating the right side.
2. JEE Main 2019: $\int e^x\left(\frac{1 - \sin x}{1 - \cos x}\right)dx = -e^x \cot(x/2) + C$. Check using differentiation.
3. Challenge: For $\int \frac{dx}{\sqrt{x} + \sqrt[4]{x}}$, one option is $\frac{4}{3}x^{3/4} - 2x^{1/2} + 4x^{1/4} - 4\ln(1+x^{1/4}) + C$. Quickly verify the leading term's coefficient by differentiation.

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Final Takeaways for Exam Day

1. First Reaction: See integration problem → immediately glance at options.
2. Quick Filter: If options contain distinct function types (log vs. inverse trig vs. polynomial), differentiation trick will be efficient.
3. Spot Check First: Before full differentiation, try $x = 0$ or $x = 1$. Often eliminates 2-3 options instantly.
4. Differentiate Systematically: Use standard derivative rules; don't rush the algebra.
5. Trust the Method: If differentiation yields the integrand (up to a constant factor you can adjust), you've found the answer.

This technique exemplifies the JEE mindset: work smarter, not just harder. It leverages the MCQ format to your advantage, turning a potential time-sink into a quick victory.

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"In the JEE, sometimes the fastest way forward is to go backwards."