The "Kuturu" Trick: Standard Linear/Quadratic Form Integration

🎯 The Strategic Insight

For integrals of the form: $I = \int \frac{dx}{ax^2 + bx + c}$ The conventional approach involves completing the square, which is time-consuming and prone to algebraic errors. The "Kuturu Trick" provides direct formulas based on the discriminant $D = b^2 - 4ac$, allowing you to write down the answer in seconds once you identify $a$, $b$, $c$, and compute $D$.

This method is exceptionally powerful for JEE because these integrals appear frequently, and the discriminant immediately tells you the form of the answer.

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📐 The Core Formulas (Memorize These)

Case 1: $D < 0$ (Complex Roots) → Arctangent Form

When the quadratic has no real roots ($4ac > b^2$), the integral yields an inverse tangent function.

$\boxed{\int \frac{dx}{ax^2 + bx + c} = \frac{2}{\sqrt{4ac - b^2}} \tan^{-1} \left( \frac{2ax + b}{\sqrt{4ac - b^2}} \right) + C}$

Memory Aid: "Negative Discriminant → Negative under root → Use Arctan".

Case 2: $D > 0$ (Distinct Real Roots) → Logarithmic Form

When the quadratic has two distinct real roots, the integral yields a logarithmic expression.

$\boxed{\int \frac{dx}{ax^2 + bx + c} = \frac{1}{\sqrt{b^2 - 4ac}} \ln \left| \frac{2ax + b - \sqrt{b^2 - 4ac}}{2ax + b + \sqrt{b^2 - 4ac}} \right| + C}$

Memory Aid: "Positive Discriminant → Positive roots → Use Log".

Case 3: $D = 0$ (Repeated Real Root) → Simple Reciprocal

When the quadratic is a perfect square:

$\int \frac{dx}{ax^2 + bx + c} = \int \frac{dx}{a(x - \alpha)^2} = -\frac{1}{a(x - \alpha)} + C$ where $\alpha = -\frac{b}{2a}$ is the repeated root.

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🔍 The Three-Step Execution

1. Identify Coefficients: Extract $a$, $b$, $c$ from $ax^2 + bx + c$.
2. Compute Discriminant: $D = b^2 - 4ac$.
3. Apply the Formula: Based on the sign of $D$, plug into the correct formula.

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📝 JEE Main Problems Solved in Seconds

Example 1: Classic Quadratic

Question: $\int \frac{dx}{x^2 - 2x + 10}$.

Kuturu Solution (30 seconds):

1. $a = 1,\ b = -2,\ c = 10$.
2. $D = (-2)^2 - 4\cdot1\cdot10 = 4 - 40 = -36 < 0$ → Use Arctan form.
3. $\sqrt{4ac - b^2} = \sqrt{40 - 4} = \sqrt{36} = 6$.
4. Apply formula: $\int \frac{dx}{x^2 - 2x + 10} = \frac{2}{6} \tan^{-1}\left( \frac{2x - 2}{6} \right) + C = \frac{1}{3} \tan^{-1}\left( \frac{x-1}{3} \right) + C$

Answer: $\boxed{\frac{1}{3} \tan^{-1}\left( \frac{x-1}{3} \right) + C}$

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Example 2: Quadratic with Real Roots

Question: $\int \frac{dx}{x^2 + x - 2}$.

Kuturu Solution (30 seconds):

1. $a = 1,\ b = 1,\ c = -2$.
2. $D = 1^2 - 4\cdot1\cdot(-2) = 1 + 8 = 9 > 0$ → Use Log form.
3. $\sqrt{D} = 3$.
4. $2ax + b = 2x + 1$.
5. Apply formula: $\int \frac{dx}{x^2 + x - 2} = \frac{1}{3} \ln \left| \frac{2x + 1 - 3}{2x + 1 + 3} \right| + C = \frac{1}{3} \ln \left| \frac{x - 1}{x + 2} \right| + C$

Answer: $\boxed{\frac{1}{3} \ln \left| \frac{x - 1}{x + 2} \right| + C}$

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Example 3: Non-Monic Quadratic (JEE Main Style)

Question: $\int \frac{dx}{2x^2 + 5x + 3}$.

Kuturu Solution (40 seconds):

1. $a = 2,\ b = 5,\ c = 3$.
2. $D = 5^2 - 4\cdot2\cdot3 = 25 - 24 = 1 > 0$ → Use Log form.
3. $\sqrt{D} = 1$.
4. $2ax + b = 4x + 5$.
5. Apply formula: $\int \frac{dx}{2x^2 + 5x + 3} = \frac{1}{1} \ln \left| \frac{4x + 5 - 1}{4x + 5 + 1} \right| + C = \ln \left| \frac{4x + 4}{4x + 6} \right| + C = \ln \left| \frac{2x + 2}{2x + 3} \right| + C$ (The factor of 2 inside the log can be absorbed into the constant $C$).

Answer: $\boxed{\ln \left| \frac{2x + 2}{2x + 3} \right| + C}$ or equivalently $\ln \left| \frac{x + 1}{2x + 3} \right| + C'$.

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🧮 Why It Works: The Derivation (For Understanding)

The trick is essentially completing the square in a disguised, pre-compiled form.

For $D < 0$: $ax^2 + bx + c = a\left[\left(x + \frac{b}{2a}\right)^2 + \frac{4ac - b^2}{4a^2}\right]$ The standard integral $\int \frac{dx}{x^2 + k^2} = \frac{1}{k} \tan^{-1}\frac{x}{k}$ applies, leading to the arctan formula.

For $D > 0$: $ax^2 + bx + c = a(x - \alpha)(x - \beta)$ Partial fractions decomposition $\frac{1}{a(x-\alpha)(x-\beta)} = \frac{1}{a(\alpha-\beta)}\left(\frac{1}{x-\alpha} - \frac{1}{x-\beta}\right)$ leads, upon integration, to the logarithmic form.

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🔄 Extended Forms and Variations

The same logic applies to related integrals. Once you recognize the pattern, you can handle variations quickly.

Variation 1: Numerator is the Derivative

$\int \frac{2ax + b}{ax^2 + bx + c} \, dx = \ln|ax^2 + bx + c| + C$ This is immediate by recognition ($u$-substitution with $u = ax^2 + bx + c$).

Variation 2: General Linear Numerator

For $\int \frac{px + q}{ax^2 + bx + c} dx$:

1. Express $px + q = \frac{p}{2a}(2ax + b) + \left(q - \frac{pb}{2a}\right)$.
2. Split the integral: $\int \frac{px+q}{ax^2+bx+c}dx = \frac{p}{2a}\int \frac{2ax+b}{ax^2+bx+c}dx + \left(q-\frac{pb}{2a}\right)\int \frac{dx}{ax^2+bx+c}$
3. The first part integrates to a log. The second part uses the Kuturu formula.

Variation 3: Integrals with Square Roots

For $\int \frac{dx}{\sqrt{ax^2 + bx + c}}$, the discriminant again dictates the form (inverse hyperbolic sin for $D<0, a>0$, inverse sin for $a<0$, and a different log form for $D>0$). The key is always to check $D$ first.

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📊 Decision Framework for JEE

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⚠️ Common Pitfalls & Safeguards

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📌 Practice for Fluency (Solve in 1 minute each)

1. $\int \frac{dx}{x^2 + 6x + 13}$
   Ans: $\frac{1}{2} \tan^{-1}\left(\frac{x+3}{2}\right) + C$

2. $\int \frac{dx}{3x^2 + 5x + 1}$
   Ans: $\frac{1}{\sqrt{13}} \ln \left| \frac{6x+5-\sqrt{13}}{6x+5+\sqrt{13}} \right| + C$

3. $\int \frac{dx}{4x^2 - 4x + 1}$ (Hint: $D=0$)
   Ans: $-\frac{1}{4x-2} + C$ or $-\frac{1}{2(2x-1)} + C$

4. JEE Main 2021 Style: If $\int \frac{\cos\theta\, d\theta}{5+7\sin\theta-2\cos^2\theta} = A\ln\|B(\theta)\|+C$, find $A$.
   Hint: Let $t=\sin\theta$, then use Kuturu trick on resulting quadratic in $t$.

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Final Strategic Summary

1. First Step is Always D: For any integral of rational functions (especially denominators), compute the discriminant immediately.
2. Sign Dictates Form: Negative D → Arctan. Positive D → Log. Zero D → Simple power rule.
3. The Formula is Reliable: It's mathematically sound, not a guess. You can derive it if needed, but in the exam, apply it directly.
4. Check Against Options: In JEE MCQs, your result might look different from an option due to algebraic simplification (e.g., $\ln|2x+4| = \ln2 + \ln|x+2|$, and $\ln2$ merges with $C$). Factor arguments completely to match.
5. Time Saving is Immense: This trick can solve in 30 seconds what takes 3+ minutes via completing the square and standard integration.

By mastering this pattern, you turn a whole class of integration problems into routine calculations, freeing up time and mental energy for the more innovative challenges in the paper.

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"In integration, recognizing the pattern is half the battle. The Kuturu trick wins that battle in one move."