The Hook: The Master Key Problem
In Mission: Impossible - Dead Reckoning (2023), Ethan Hunt needs to unlock a series of complex locks. Some locks need a simple key (direct approach), but others need special tools and multi-step techniques.
Similarly, in integration:
- Simple lock: $\int x^2 \, dx$ → Direct formula (basic key)
- Complex lock: $\int x \cdot e^x \, dx$ → Needs integration by parts (special tool)
- Tricky lock: $\int \frac{2x+3}{x^2-5x+6} \, dx$ → Needs partial fractions (multi-step technique)
JEE Reality: About 60% of integration problems need these advanced techniques. This is a must-master topic for both JEE Main and Advanced!
Technique 1: Integration by Substitution
The Core Concept
Substitution is the reverse of the chain rule. If you can spot a function and its derivative hiding in the integrand, substitution unlocks the problem.
$$\boxed{\int f(g(x)) \cdot g'(x) \, dx = \int f(u) \, du \quad \text{where } u = g(x)}$$In simple terms: Replace a complicated part with a simple variable $u$, integrate, then substitute back.
When to Use Substitution
Look for these patterns:
$f(x) \cdot f'(x)$ pattern: $\int x \cos(x^2) \, dx$ → $u = x^2$, $du = 2x \, dx$
$\frac{f'(x)}{f(x)}$ pattern: $\int \frac{2x}{x^2+1} \, dx$ → Gives $\ln|f(x)| + C$
$f'(x) \cdot [f(x)]^n$ pattern: $\int 2x(x^2+1)^5 \, dx$ → $u = x^2 + 1$
Composite function: $\int \sin(3x) \, dx$ → $u = 3x$
Quick test: Can I find a function whose derivative appears as a factor?
Step-by-Step Process
Step 1: Identify the substitution $u = g(x)$ Step 2: Find $du = g'(x) \, dx$ Step 3: Replace all $x$ terms with $u$ terms Step 4: Integrate with respect to $u$ Step 5: Substitute back to get the answer in terms of $x$
Worked Examples
Problem: Evaluate $\int 2x \cos(x^2) \, dx$
Solution:
Step 1: Let $u = x^2$ (inner function)
Step 2: Then $du = 2x \, dx$ (perfect! we have $2x \, dx$ in the integral)
Step 3: Rewrite:
$$\int 2x \cos(x^2) \, dx = \int \cos(u) \, du$$Step 4: Integrate:
$$\int \cos(u) \, du = \sin(u) + C$$Step 5: Substitute back:
$$= \sin(x^2) + C$$Verification: $\frac{d}{dx}[\sin(x^2)] = \cos(x^2) \cdot 2x$ ✓
Problem: Find $\int \frac{2x + 3}{x^2 + 3x + 5} \, dx$
Solution:
Notice: If $f(x) = x^2 + 3x + 5$, then $f'(x) = 2x + 3$ (numerator!)
Special rule: $\int \frac{f'(x)}{f(x)} \, dx = \ln|f(x)| + C$
$$\int \frac{2x + 3}{x^2 + 3x + 5} \, dx = \ln|x^2 + 3x + 5| + C$$JEE Shortcut: Always check if the numerator is the derivative of the denominator!
Problem: Evaluate $\int \frac{x}{x^2 + 3x + 5} \, dx$
Solution:
Numerator is $x$, but derivative of denominator is $2x + 3$.
Strategy: Write $x = \frac{1}{2}(2x) = \frac{1}{2}(2x + 3 - 3)$
$$\int \frac{x}{x^2 + 3x + 5} \, dx = \int \frac{\frac{1}{2}(2x + 3) - \frac{3}{2}}{x^2 + 3x + 5} \, dx$$ $$= \frac{1}{2}\int \frac{2x + 3}{x^2 + 3x + 5} \, dx - \frac{3}{2}\int \frac{1}{x^2 + 3x + 5} \, dx$$ $$= \frac{1}{2}\ln|x^2 + 3x + 5| - \frac{3}{2} \cdot \text{(requires completing the square)}$$Key skill: Sometimes you need to manipulate the numerator to match the derivative!
Problem: Evaluate $\int \sin^3 x \cos x \, dx$
Solution:
Notice: $\frac{d}{dx}(\sin x) = \cos x$ (the derivative is present!)
Let $u = \sin x$, then $du = \cos x \, dx$
$$\int \sin^3 x \cos x \, dx = \int u^3 \, du = \frac{u^4}{4} + C = \frac{\sin^4 x}{4} + C$$Pattern: When you see $\sin^n x \cos x$ or $\cos^n x \sin x$, use substitution!
Technique 2: Integration by Parts
The Core Concept
Integration by parts is the reverse of the product rule for differentiation.
$$\boxed{\int u \, dv = uv - \int v \, du}$$Or in expanded form:
$$\boxed{\int u(x) \cdot v'(x) \, dx = u(x) \cdot v(x) - \int v(x) \cdot u'(x) \, dx}$$In simple terms: To integrate a product, sacrifice one function (differentiate it) and integrate the other, then deal with what’s left over.
The LIATE Rule (Memory Trick)
When deciding which function to choose as $u$ (to differentiate), use the LIATE priority:
L - Logarithmic functions ($\ln x$, $\log x$) I - Inverse trigonometric ($\sin^{-1} x$, $\tan^{-1} x$, etc.) A - Algebraic ($x$, $x^2$, $x^n$) T - Trigonometric ($\sin x$, $\cos x$, $\tan x$) E - Exponential ($e^x$, $a^x$)
Choose $u$ as the function highest on the list, and $dv$ as the remaining part.
Why?
- Logarithmic functions become simpler when differentiated: $(\ln x)' = \frac{1}{x}$
- Exponential functions stay the same: $(e^x)' = e^x$, so better to integrate them
Example: In $\int x e^x \, dx$:
- $x$ is Algebraic (A)
- $e^x$ is Exponential (E)
- Since A comes before E in LIATE, choose $u = x$, $dv = e^x \, dx$
Step-by-Step Process
Step 1: Choose $u$ and $dv$ using LIATE Step 2: Find $du$ by differentiating $u$ Step 3: Find $v$ by integrating $dv$ Step 4: Apply formula: $\int u \, dv = uv - \int v \, du$ Step 5: Evaluate the remaining integral
Interactive Demo: Visualize Integration Techniques
Explore how different integration methods transform functions interactively.
Worked Examples
Problem: Evaluate $\int x e^x \, dx$
Solution:
Using LIATE: $x$ (Algebraic) comes before $e^x$ (Exponential)
- Let $u = x$ → $du = dx$
- Let $dv = e^x \, dx$ → $v = e^x$
Apply formula:
$$\int x e^x \, dx = x \cdot e^x - \int e^x \cdot 1 \, dx$$ $$= x e^x - e^x + C = e^x(x - 1) + C$$Verification: $\frac{d}{dx}[e^x(x-1)] = e^x(x-1) + e^x \cdot 1 = xe^x$ ✓
Problem: Find $\int \ln x \, dx$
Solution:
This looks like a single function, but we can write it as $\int \ln x \cdot 1 \, dx$
Using LIATE: $\ln x$ (Logarithmic) comes before $1$ (constant, treat as Algebraic)
- Let $u = \ln x$ → $du = \frac{1}{x} \, dx$
- Let $dv = 1 \, dx$ → $v = x$
Apply formula:
$$\int \ln x \, dx = (\ln x) \cdot x - \int x \cdot \frac{1}{x} \, dx$$ $$= x\ln x - \int 1 \, dx = x\ln x - x + C$$ $$= x(\ln x - 1) + C$$JEE Pattern: Logarithmic and inverse trig functions often need this “multiply by 1” trick!
Problem: Evaluate $\int x^2 e^x \, dx$
Solution:
First application:
- $u = x^2$ → $du = 2x \, dx$
- $dv = e^x \, dx$ → $v = e^x$
Second application (on $\int x e^x \, dx$):
- $u = x$ → $du = dx$
- $dv = e^x \, dx$ → $v = e^x$
Combine:
$$\int x^2 e^x \, dx = x^2 e^x - 2(xe^x - e^x) + C$$ $$= x^2 e^x - 2xe^x + 2e^x + C = e^x(x^2 - 2x + 2) + C$$Pattern: For $\int x^n e^x \, dx$, you need to apply by parts $n$ times!
Problem: Find $\int e^x \sin x \, dx$
Solution:
Let $I = \int e^x \sin x \, dx$
First application:
- $u = \sin x$ → $du = \cos x \, dx$
- $dv = e^x \, dx$ → $v = e^x$
Second application (on $\int e^x \cos x \, dx$):
- $u = \cos x$ → $du = -\sin x \, dx$
- $dv = e^x \, dx$ → $v = e^x$
Substitute back:
$$I = e^x \sin x - (e^x \cos x + I)$$ $$I = e^x \sin x - e^x \cos x - I$$ $$2I = e^x(\sin x - \cos x)$$ $$I = \frac{e^x(\sin x - \cos x)}{2} + C$$Magic trick: The integral appeared on both sides! This “circular” technique works for $\int e^{ax} \sin bx \, dx$ and $\int e^{ax} \cos bx \, dx$.
Technique 3: Partial Fractions
The Core Concept
Partial fractions is used when integrating rational functions (ratio of polynomials) where the degree of the numerator is less than the degree of the denominator.
Strategy: Break a complex fraction into simpler fractions that we can integrate easily.
$$\frac{P(x)}{Q(x)} = \frac{A}{x - a} + \frac{B}{x - b} + \ldots$$When to Use Partial Fractions
Given: $\int \frac{P(x)}{Q(x)} \, dx$ (ratio of polynomials)
Step 1: Is degree of $P(x) < $ degree of $Q(x)$?
- No: Do polynomial long division first
- Yes: Proceed to factorization
Step 2: Can you factor $Q(x)$ into linear and/or quadratic factors?
- Yes: Use partial fractions
- No: May need advanced techniques (substitution, completing the square)
Step 3: Check if numerator is related to derivative of denominator
- If $P(x) = Q'(x)$ → Direct formula: $\ln|Q(x)| + C$
Types of Partial Fractions
Type 1: Distinct Linear Factors
$$\frac{P(x)}{(x-a)(x-b)} = \frac{A}{x-a} + \frac{B}{x-b}$$Problem: Evaluate $\int \frac{1}{x^2 - 5x + 6} \, dx$
Solution:
Step 1: Factor the denominator:
$$x^2 - 5x + 6 = (x-2)(x-3)$$Step 2: Set up partial fractions:
$$\frac{1}{(x-2)(x-3)} = \frac{A}{x-2} + \frac{B}{x-3}$$Step 3: Multiply both sides by $(x-2)(x-3)$:
$$1 = A(x-3) + B(x-2)$$Step 4: Find $A$ and $B$:
Method 1 (Substitution):
- Let $x = 2$: $1 = A(-1) + 0$ → $A = -1$
- Let $x = 3$: $1 = 0 + B(1)$ → $B = 1$
Method 2 (Coefficient comparison):
- $1 = Ax - 3A + Bx - 2B = (A+B)x + (-3A-2B)$
- Coefficient of $x$: $0 = A + B$
- Constant term: $1 = -3A - 2B$
- Solving: $A = -1$, $B = 1$
Step 5: Integrate:
$$\int \frac{1}{x^2 - 5x + 6} \, dx = \int \left(\frac{-1}{x-2} + \frac{1}{x-3}\right) dx$$ $$= -\ln|x-2| + \ln|x-3| + C = \ln\left|\frac{x-3}{x-2}\right| + C$$Type 2: Repeated Linear Factors
$$\frac{P(x)}{(x-a)^n} = \frac{A_1}{x-a} + \frac{A_2}{(x-a)^2} + \ldots + \frac{A_n}{(x-a)^n}$$Problem: Find $\int \frac{2x + 1}{(x-1)^2} \, dx$
Solution:
Step 1: Set up partial fractions:
$$\frac{2x + 1}{(x-1)^2} = \frac{A}{x-1} + \frac{B}{(x-1)^2}$$Step 2: Multiply by $(x-1)^2$:
$$2x + 1 = A(x-1) + B$$Step 3: Find constants:
- Let $x = 1$: $3 = 0 + B$ → $B = 3$
- Coefficient of $x$: $2 = A$ → $A = 2$
Step 4: Integrate:
$$\int \frac{2x + 1}{(x-1)^2} \, dx = \int \left(\frac{2}{x-1} + \frac{3}{(x-1)^2}\right) dx$$ $$= 2\ln|x-1| + 3 \cdot \frac{(x-1)^{-1}}{-1} + C$$ $$= 2\ln|x-1| - \frac{3}{x-1} + C$$Type 3: Irreducible Quadratic Factors
$$\frac{P(x)}{(x^2 + bx + c)} = \frac{Ax + B}{x^2 + bx + c}$$Problem: Evaluate $\int \frac{3x + 5}{x^2 + 2x + 5} \, dx$
Solution:
Step 1: Check if numerator is related to derivative of denominator.
- $\frac{d}{dx}(x^2 + 2x + 5) = 2x + 2$
Step 2: Write $3x + 5 = \frac{3}{2}(2x + 2) + (5 - 3) = \frac{3}{2}(2x + 2) + 2$
Step 3: Split the integral:
$$\int \frac{3x + 5}{x^2 + 2x + 5} \, dx = \int \frac{\frac{3}{2}(2x + 2)}{x^2 + 2x + 5} \, dx + \int \frac{2}{x^2 + 2x + 5} \, dx$$ $$= \frac{3}{2}\ln|x^2 + 2x + 5| + 2\int \frac{1}{x^2 + 2x + 5} \, dx$$Step 4: For the second integral, complete the square:
$$x^2 + 2x + 5 = (x+1)^2 + 4$$ $$\int \frac{1}{(x+1)^2 + 4} \, dx = \frac{1}{2}\tan^{-1}\left(\frac{x+1}{2}\right) + C$$Final answer:
$$= \frac{3}{2}\ln|x^2 + 2x + 5| + 2 \cdot \frac{1}{2}\tan^{-1}\left(\frac{x+1}{2}\right) + C$$ $$= \frac{3}{2}\ln|x^2 + 2x + 5| + \tan^{-1}\left(\frac{x+1}{2}\right) + C$$Memory Tricks & Patterns
Integration Decision Flowchart
1. Can you use a standard formula? → Do it! (Fastest method)
2. Is it a product of two functions?
- Is one function’s derivative present? → Substitution
- Not related by derivatives? → Integration by parts (use LIATE)
3. Is it a rational function (fraction)?
- Numerator = derivative of denominator? → $\ln$ formula
- Not quite? → Adjust numerator or use partial fractions
4. Is it a trigonometric integral? → See Trigonometric Integration
5. Still stuck? → Try algebraic manipulation, trig identities, or advanced substitutions
LIATE Mnemonic
"Lately I Ate Two Eggs"
- Lately = Logarithmic
- I = Inverse trig
- Ate = Algebraic
- Two = Trigonometric
- Eggs = Exponential
Common Mistakes to Avoid
WRONG: In $\int x e^x \, dx$, choosing $u = e^x$ and $dv = x \, dx$
This gives: $\int x e^x \, dx = e^x \cdot \frac{x^2}{2} - \int \frac{x^2}{2} e^x \, dx$
The second integral is more complicated than the original!
CORRECT: Use LIATE—choose $u = x$ (simpler after differentiation)
Problem: $\int x \sin(x^2) \, dx$
WRONG: Let $u = x^2$, $du = 2x \, dx$, then $\int \sin u \, du$ (forgot to adjust!)
CORRECT: Since $du = 2x \, dx$, we have $x \, dx = \frac{du}{2}$
$$\int x \sin(x^2) \, dx = \int \sin u \cdot \frac{du}{2} = \frac{1}{2}\int \sin u \, du = -\frac{1}{2}\cos u + C = -\frac{1}{2}\cos(x^2) + C$$WRONG: Trying to decompose $\frac{1}{x^2 + 1}$ into $\frac{A}{x} + \frac{B}{x}$ (doesn’t factor!)
CORRECT: $x^2 + 1$ is irreducible over real numbers. Use the standard formula:
$$\int \frac{1}{x^2 + 1} \, dx = \tan^{-1} x + C$$WRONG: $\int \frac{1}{x-2} \, dx = \ln(x-2) + C$
CORRECT: $\int \frac{1}{x-2} \, dx = \ln|x-2| + C$
The absolute value is crucial for $x < 2$!
Practice Problems
Level 1: Foundation (NCERT)
Evaluate: $\int \frac{1}{2x + 3} \, dx$
Solution: Let $u = 2x + 3$, then $du = 2 \, dx$, so $dx = \frac{du}{2}$
$$\int \frac{1}{2x + 3} \, dx = \int \frac{1}{u} \cdot \frac{du}{2} = \frac{1}{2}\ln|u| + C = \frac{1}{2}\ln|2x+3| + C$$Find: $\int x \sin x \, dx$
Solution: Using LIATE: $u = x$ (Algebraic), $dv = \sin x \, dx$ (Trigonometric)
- $u = x$ → $du = dx$
- $dv = \sin x \, dx$ → $v = -\cos x$
Evaluate: $\int \frac{1}{x(x+1)} \, dx$
Solution:
$$\frac{1}{x(x+1)} = \frac{A}{x} + \frac{B}{x+1}$$$1 = A(x+1) + Bx$
- Let $x = 0$: $1 = A$ → $A = 1$
- Let $x = -1$: $1 = -B$ → $B = -1$
Level 2: JEE Main
JEE Main Pattern: Find $\int \frac{x^2}{(x^3 + 1)^4} \, dx$
Solution: Let $u = x^3 + 1$, then $du = 3x^2 \, dx$, so $x^2 \, dx = \frac{du}{3}$
$$\int \frac{x^2}{(x^3 + 1)^4} \, dx = \int \frac{1}{u^4} \cdot \frac{du}{3} = \frac{1}{3}\int u^{-4} \, du$$ $$= \frac{1}{3} \cdot \frac{u^{-3}}{-3} + C = -\frac{1}{9u^3} + C = -\frac{1}{9(x^3+1)^3} + C$$Problem: Evaluate $\int (2x + 3)e^x \, dx$
Solution: Using LIATE: $u = 2x + 3$, $dv = e^x \, dx$
- $u = 2x + 3$ → $du = 2 \, dx$
- $dv = e^x \, dx$ → $v = e^x$
Problem: Find $\int \frac{5x - 2}{x^2 - 4} \, dx$
Solution: Factor: $x^2 - 4 = (x-2)(x+2)$
Partial fractions:
$$\frac{5x - 2}{(x-2)(x+2)} = \frac{A}{x-2} + \frac{B}{x+2}$$$5x - 2 = A(x+2) + B(x-2)$
- Let $x = 2$: $8 = 4A$ → $A = 2$
- Let $x = -2$: $-12 = -4B$ → $B = 3$
Level 3: JEE Advanced
JEE Advanced Pattern: Evaluate $\int x^2 \ln x \, dx$
Solution: Using LIATE: $u = \ln x$ (Logarithmic), $dv = x^2 \, dx$ (Algebraic)
- $u = \ln x$ → $du = \frac{1}{x} \, dx$
- $dv = x^2 \, dx$ → $v = \frac{x^3}{3}$
Problem: Find $\int \frac{e^x}{e^x + 1} \, dx$
Solution: Notice: If $u = e^x + 1$, then $du = e^x \, dx$ (perfect!)
$$\int \frac{e^x}{e^x + 1} \, dx = \int \frac{1}{u} \, du = \ln|u| + C = \ln|e^x + 1| + C$$Since $e^x + 1 > 0$ for all $x$:
$$= \ln(e^x + 1) + C$$Alternative approach: $\frac{e^x}{e^x + 1} = 1 - \frac{1}{e^x + 1}$ also works!
Conceptual Problem: If $I_n = \int x^n e^x \, dx$, prove that $I_n = x^n e^x - n I_{n-1}$
Solution: Use integration by parts with $u = x^n$, $dv = e^x \, dx$:
- $u = x^n$ → $du = nx^{n-1} \, dx$
- $dv = e^x \, dx$ → $v = e^x$
Application: This is a reduction formula—allows you to compute $I_n$ in terms of $I_{n-1}$, eventually reaching $I_0 = e^x$.
This pattern is crucial for JEE Advanced!
Quick Revision Box
| Method | When to Use | Example |
|---|---|---|
| Substitution | $f(g(x)) \cdot g'(x)$ pattern | $\int x\cos(x^2) dx$ |
| $\frac{f'(x)}{f(x)}$ | Numerator = derivative of denominator | $\int \frac{2x}{x^2+1} dx = \ln(x^2+1)$ |
| By Parts (LIATE) | Product of unrelated functions | $\int x e^x dx$ |
| Partial Fractions | Rational function (proper) | $\int \frac{1}{x^2-1} dx$ |
| Completing Square | Quadratic in denominator | $\int \frac{1}{x^2+4x+5} dx$ |
Cross-Links & Prerequisites
Before studying this topic:
- Indefinite Integrals - Standard formulas
- Differentiation Rules - Chain rule, product rule
- Higher Derivatives - For repeated integration by parts
Related topics:
- Trigonometric Integration - Special methods for trig functions
- Definite Integrals - Applying these techniques with limits
- Inverse Trigonometry - For inverse trig substitutions
Applications:
- Differential Equations - Solving ODEs requires integration techniques
- Variable Separable Method - Direct application
Algebraic Prerequisites:
- Complex Numbers - For partial fractions with complex roots
Teacher’s Summary
Substitution = Reverse Chain Rule: Look for a function and its derivative hiding in the integrand. Master the $\frac{f'(x)}{f(x)} = \ln|f(x)|$ pattern.
LIATE is Your Best Friend: For integration by parts, always use LIATE to choose $u$. Logarithmic and inverse trig functions should almost always be $u$.
Partial Fractions: Factor First: You can’t decompose what you can’t factor. Linear factors give logarithms, quadratic factors need completing the square.
Practice Decision Making: The hardest part isn’t doing the integration—it’s choosing the right method. Build your pattern recognition by solving 50+ problems.
Verify by Differentiation: After finding an antiderivative, always check by differentiating. This catches silly mistakes and builds confidence.
“Integration is 50% technique and 50% pattern recognition. Master the techniques here, and you’ll solve 80% of JEE integration problems.”
JEE Strategy: In the exam, spend 10-15 seconds deciding the method before starting. A wrong method can waste 3-4 minutes!
What’s Next?
You’ve mastered the core techniques! Now specialize:
- Next: Trigonometric Integration - Special identities and transformations for trig integrals
- Then: Definite Integrals - Adding limits and using properties
- Finally: Applications of Integrals - Finding areas, volumes, and solving real problems
Keep practicing—integration mastery is a marathon, not a sprint!