Cracking Irrational Integrals in JEE: Smart Substitutions
Master Euler substitutions, trigonometric tricks, and special methods to solve any $\sqrt{ax^2 + bx + c}$ integral in seconds.
Why Irrational Integrals Matter in JEE
Integrals containing $\sqrt{ax^2 + bx + c}$ appear in every JEE Advanced paper and are worth 4-8 marks. Most students struggle because they don't know which substitution to use when.
🎯 JEE Exam Pattern Insight
- 1-2 questions per paper on irrational integrals
- Often combined with definite integrals and properties
- Can be solved in 30-60 seconds with the right method
- High discrimination value - separates top rankers
🚀 Method Navigation
1. Euler's First Substitution
When $a > 0$ in $\sqrt{ax^2 + bx + c}$, use:
Substitution Formula
Squares both sides and eliminates the square root!
Example: Solve $\int \frac{dx}{\sqrt{x^2 + 4x + 6}}$
Step-by-Step Solution:
Step 1: Complete the square: $x^2 + 4x + 6 = (x+2)^2 + 2$
Step 2: Apply Euler's 1st substitution: $\sqrt{(x+2)^2 + 2} = t - (x+2)$
Step 3: Square both sides: $(x+2)^2 + 2 = t^2 - 2t(x+2) + (x+2)^2$
Step 4: Simplify: $2 = t^2 - 2t(x+2) \Rightarrow x+2 = \frac{t^2 - 2}{2t}$
Step 5: Differentiate: $dx = \frac{t^2 + 2}{2t^2} dt$
Step 6: Substitute back: $\int \frac{dx}{\sqrt{(x+2)^2 + 2}} = \int \frac{1}{t - (x+2)} dx$
Step 7: After substitution: $\int \frac{dt}{t} = \ln|t| + C$
Step 8: Back substitute: $t = \sqrt{(x+2)^2 + 2} + (x+2)$
Final Answer:
$\ln|\sqrt{x^2 + 4x + 6} + x + 2| + C$
2. Euler's Second Substitution
When $c > 0$ in $\sqrt{ax^2 + bx + c}$, use:
Substitution Formula
Perfect when the constant term is positive!
Example: Solve $\int \frac{dx}{x\sqrt{x^2 + x + 1}}$
Step-by-Step Solution:
Step 1: Apply Euler's 2nd substitution: $\sqrt{x^2 + x + 1} = xt + 1$
Step 2: Square both sides: $x^2 + x + 1 = x^2t^2 + 2xt + 1$
Step 3: Simplify: $x + 1 = xt^2 + 2t \Rightarrow x = \frac{1 - 2t}{t^2 - 1}$
Step 4: Differentiate: $dx = \frac{2t^2 - 2t + 2}{(t^2 - 1)^2} dt$
Step 5: Original integral becomes: $\int \frac{1}{x(xt+1)} dx$
Step 6: After substitution: $\int \frac{t^2 - 1}{1 - 2t} \cdot \frac{1}{\frac{1 - 2t}{t^2 - 1}t + 1} \cdot \frac{2t^2 - 2t + 2}{(t^2 - 1)^2} dt$
Step 7: Simplify dramatically to: $\int -2 dt = -2t + C$
Step 8: Back substitute: $t = \frac{\sqrt{x^2 + x + 1} - 1}{x}$
Final Answer:
$-2\cdot\frac{\sqrt{x^2 + x + 1} - 1}{x} + C$
3. Euler's Third Substitution
When the quadratic has real roots $α$ and $β$, use:
Substitution Formula
Most powerful when you can factor the quadratic!
Example: Solve $\int \frac{dx}{\sqrt{x^2 - 5x + 6}}$
Step-by-Step Solution:
Step 1: Factor: $x^2 - 5x + 6 = (x-2)(x-3)$
Step 2: Apply Euler's 3rd substitution: $\sqrt{(x-2)(x-3)} = t(x-2)$
Step 3: Square both sides: $(x-2)(x-3) = t^2(x-2)^2$
Step 4: Cancel $(x-2)$: $x-3 = t^2(x-2)$
Step 5: Solve for x: $x = \frac{3 - 2t^2}{1 - t^2}$
Step 6: Differentiate: $dx = \frac{2t}{(1-t^2)^2} dt$
Step 7: Original integral becomes: $\int \frac{1}{t(x-2)} dx$
Step 8: After substitution: $\int \frac{1}{t} \cdot \frac{2t}{(1-t^2)^2} dt = 2\int \frac{dt}{(1-t^2)^2}$
Step 9: Use partial fractions and integrate
4. Trigonometric Substitutions
The fastest method for standard forms - memorize these!
For $\sqrt{a^2 - x^2}$
$x = a\sin\theta$
$\sqrt{a^2 - x^2} = a\cos\theta$
For $\sqrt{a^2 + x^2}$
$x = a\tan\theta$
$\sqrt{a^2 + x^2} = a\sec\theta$
For $\sqrt{x^2 - a^2}$
$x = a\sec\theta$
$\sqrt{x^2 - a^2} = a\tan\theta$
Universal Sub
$x = a\cos 2\theta$
Works for many forms
Example: Solve $\int \sqrt{4 - x^2} dx$
Lightning Fast Solution:
Step 1: $x = 2\sin\theta \Rightarrow dx = 2\cos\theta d\theta$
Step 2: $\sqrt{4 - x^2} = \sqrt{4 - 4\sin^2\theta} = 2\cos\theta$
Step 3: Integral becomes: $\int 2\cos\theta \cdot 2\cos\theta d\theta = 4\int \cos^2\theta d\theta$
Step 4: $4\int \frac{1 + \cos 2\theta}{2} d\theta = 2\int (1 + \cos 2\theta) d\theta$
Step 5: $2(\theta + \frac{\sin 2\theta}{2}) + C = 2\theta + \sin 2\theta + C$
Step 6: Back substitute: $\theta = \sin^{-1}(\frac{x}{2})$
$\sin 2\theta = 2\sin\theta\cos\theta = 2\cdot\frac{x}{2}\cdot\frac{\sqrt{4-x^2}}{2} = \frac{x\sqrt{4-x^2}}{2}$
Final Answer:
$2\sin^{-1}\left(\frac{x}{2}\right) + \frac{x\sqrt{4-x^2}}{2} + C$
5. Special Forms & Direct Formulas
Memorize these - they save precious minutes in JEE!
Standard Form 1
$\int \frac{dx}{\sqrt{x^2 + a^2}} = \ln|x + \sqrt{x^2 + a^2}| + C$
Standard Form 2
$\int \frac{dx}{\sqrt{x^2 - a^2}} = \ln|x + \sqrt{x^2 - a^2}| + C$
Standard Form 3
$\int \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}\left(\frac{x}{a}\right) + C$
Reduction Formula
$\int \sqrt{x^2 + a^2} dx = \frac{x}{2}\sqrt{x^2 + a^2} + \frac{a^2}{2}\ln|x + \sqrt{x^2 + a^2}| + C$
🎯 Quick Decision Guide: Which Method When?
| Integral Form | Best Method | Time Saved |
|---|---|---|
| $\sqrt{ax^2 + bx + c}$, $a > 0$ | Euler's 1st | 2-3 minutes |
| $\sqrt{ax^2 + bx + c}$, $c > 0$ | Euler's 2nd | 1-2 minutes |
| $\sqrt{(x-α)(x-β)}$ | Euler's 3rd | 3-4 minutes |
| $\sqrt{a^2 ± x^2}$, $\sqrt{x^2 - a^2}$ | Trigonometric | 1-2 minutes |
| Standard forms | Direct formula | 30-60 seconds |
📝 Practice These JEE Problems
1. $\int \frac{dx}{\sqrt{2x^2 + 3x + 4}}$
2. $\int \frac{dx}{x\sqrt{x^2 + x + 1}}$
3. $\int \sqrt{x^2 + 4x + 5} dx$
4. $\int \frac{x dx}{\sqrt{1 - x^2}}$
🚀 Exam Day Quick Tips
Time-Saving Strategies
- Always complete the square first - reveals the best method
- Memorize the 3 Euler substitutions and when to use them
- Know trigonometric substitutions for standard forms
- Practice direct formulas for common integrals
Common Pitfalls to Avoid
- Forgetting $+C$ in indefinite integrals (-1 mark)
- Missing absolute values in logarithms
- Wrong domain in inverse trigonometric functions
- Algebraic errors during substitution
Master Irrational Integrals for JEE 2026!
These techniques can help you secure 4-8 marks in every JEE paper