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Integral Calculus · Module 1 of 5 · JEE Main & Advanced
Standard Formulas & Integration by Substitution
Module 1 Objective
This module establishes the entire foundation of integration. Every advanced technique relies on the
standard formulas memorised here. No student should proceed to later modules without mastering
every row of the two tables below.
Chapter 1: The Anti-Derivative — Core Idea
If \(\dfrac{d}{dx}[F(x)] = f(x)\), then we define the indefinite integral:
\[\int f(x)\,dx = F(x) + C\]
The constant \(C\) (constant of integration) is mandatory — the integral of a function represents an
entire family of curves, not one unique curve. Omitting \(C\) in JEE = guaranteed mark
loss.
Why C Matters
Both \(F(x) = x^2\) and \(G(x) = x^2 + 7\) have the same derivative \(2x\). So \(\int
2x\,dx\) correctly represents both — written as \(x^2 + C\). The constant captures the entire
family.
Chapter 2: Standard Integral Formulas — The Arsenal
These are non-negotiable memorisation targets. Every JEE question (even hard ones) bottoms out at one of
these.
Table A — Algebraic & Exponential
| # |
Function \(f(x)\) |
\(\int f(x)\,dx\) |
Note |
| 1 |
\(x^n\) \((n \ne -1)\) |
\(\dfrac{x^{n+1}}{n+1} + C\) |
Power Rule — works for any real \(n\) |
| 2 |
\(\dfrac{1}{x}\) |
\(\ln|x| + C\) |
The \(n=-1\) exception. The modulus is mandatory. |
| 3 |
\(e^x\) |
\(e^x + C\) |
Only function equal to its own integral! |
| 4 |
\(a^x\) |
\(\dfrac{a^x}{\ln a} + C\) |
Base \(a > 0, a \ne 1\) |
| 5 |
\(\sqrt{x}\) |
\(\dfrac{2}{3}x^{3/2} + C\) |
Special case of rule 1 with \(n=1/2\) |
| 6 |
\(\dfrac{1}{\sqrt{x}}\) |
\(2\sqrt{x} + C\) |
\(n = -1/2\) |
Table B — Trigonometric
| # |
Function \(f(x)\) |
\(\int f(x)\,dx\) |
Memory Key |
| 7 |
\(\sin x\) |
\(-\cos x + C\) |
Integral has a minus |
| 8 |
\(\cos x\) |
\(\sin x + C\) |
No minus here |
| 9 |
\(\tan x\) |
\(\ln|\sec x| + C\) |
Also written \(-\ln|\cos x|+C\) |
| 10 |
\(\cot x\) |
\(\ln|\sin x| + C\) |
|
| 11 |
\(\sec x\) |
\(\ln|\sec x + \tan x| + C\) |
Multiply trick in derivation |
| 12 |
\(\csc x\) |
\(\ln|\csc x - \cot x| + C\) |
|
| 13 |
\(\sec^2 x\) |
\(\tan x + C\) |
Reverse of \(\frac{d}{dx}[\tan x]\) |
| 14 |
\(\csc^2 x\) |
\(-\cot x + C\) |
Minus sign! |
| 15 |
\(\sec x \tan x\) |
\(\sec x + C\) |
|
| 16 |
\(\csc x \cot x\) |
\(-\csc x + C\) |
|
Table C — Inverse Trigonometric & Logarithmic Forms HIGH YIELD
| # |
Function \(f(x)\) |
\(\int f(x)\,dx\) |
| 17 |
\(\dfrac{1}{\sqrt{1-x^2}}\) |
\(\sin^{-1}x + C\) |
| 18 |
\(\dfrac{-1}{\sqrt{1-x^2}}\) |
\(\cos^{-1}x + C\) |
| 19 |
\(\dfrac{1}{1+x^2}\) |
\(\tan^{-1}x + C\) |
| 20 |
\(\dfrac{1}{a^2+x^2}\) |
\(\dfrac{1}{a}\tan^{-1}\!\dfrac{x}{a} + C\) |
| 21 |
\(\dfrac{1}{\sqrt{a^2-x^2}}\) |
\(\sin^{-1}\!\dfrac{x}{a} + C\) |
| 22 |
\(\dfrac{1}{x^2-a^2}\) |
\(\dfrac{1}{2a}\ln\!\left|\dfrac{x-a}{x+a}\right| + C\) |
| 23 |
\(\dfrac{1}{a^2-x^2}\) |
\(\dfrac{1}{2a}\ln\!\left|\dfrac{a+x}{a-x}\right| + C\) |
| 24 |
\(\dfrac{1}{\sqrt{x^2\pm a^2}}\) |
\(\ln\!\left|x+\sqrt{x^2\pm a^2}\right| + C\) |
| 25 |
\(\sqrt{a^2-x^2}\) |
\(\dfrac{x}{2}\sqrt{a^2-x^2}+\dfrac{a^2}{2}\sin^{-1}\!\dfrac{x}{a}+C\) |
| 26 |
\(\sqrt{x^2\pm a^2}\) |
\(\dfrac{x}{2}\sqrt{x^2\pm a^2}\pm\dfrac{a^2}{2}\ln\!\left|x+\sqrt{x^2\pm a^2}\right|+C\) |
Chapter 3: Immediate Deductions — Shifting the Argument
All formulas above work for expressions of the form \(f(ax+b)\). Just divide by the inner derivative \(a\):
Chain Rule Reverse
\[\int f(ax+b)\,dx = \frac{1}{a}F(ax+b) + C\]
Examples: \(\int e^{3x}\,dx = \dfrac{e^{3x}}{3}+C\)
| \(\int \sin(2x)\,dx = -\dfrac{\cos 2x}{2}+C\) | \(\int(3x+1)^5\,dx =
\dfrac{(3x+1)^6}{18}+C\)
Chapter 4: Integration by Substitution — The Master Method
When the integrand contains a function and its derivative, a substitution converts it into a
standard form. This is the most frequently used method in JEE.
Core Technique
Step 1: Identify the "inside" function \(g(x)\) whose derivative \(g'(x)\) also
appears.
Step 2: Set \(t = g(x)\), giving \(dt = g'(x)\,dx\).
Step 3: The integral becomes \(\int f(t)\,dt\) — a standard form.
Step 4: Integrate and substitute back \(t = g(x)\).
Pattern Recognition Guide
| Integrand Pattern |
Substitution |
Result Form |
| \([f(x)]^n \cdot f'(x)\) |
\(t = f(x)\) |
\(\dfrac{[f(x)]^{n+1}}{n+1}+C\) |
| \(\dfrac{f'(x)}{f(x)}\) |
\(t = f(x)\) |
\(\ln|f(x)|+C\) |
| \(f'(x) \cdot e^{f(x)}\) |
\(t = f(x)\) |
\(e^{f(x)}+C\) |
| \(\dfrac{f'(x)}{\sqrt{f(x)}}\) |
\(t = f(x)\) |
\(2\sqrt{f(x)}+C\) |
Q1. Evaluate \(\displaystyle\int \frac{2x}{x^2+1}\,dx\)
Solution:
Numerator 2x = derivative of denominator \(x^2+1\). The integral is of type \(\dfrac{f'(x)}{f(x)}\).
Let \(t = x^2+1\), \(dt = 2x\,dx\). \(\displaystyle\int\frac{dt}{t} = \ln|t|+C = \boxed{\ln(x^2+1)+C}\)
Q2. Evaluate \(\displaystyle\int \frac{\cos(\ln x)}{x}\,dx\)
Solution:
Let \(t=\ln x\), \(dt = \dfrac{dx}{x}\). The integral transforms to \(\displaystyle\int\cos t\,dt = \sin t +
C = \boxed{\sin(\ln x)+C}\)
Q3. Evaluate \(\displaystyle\int \frac{x^3}{(x^4+1)^3}\,dx\)
Solution:
Let \(t=x^4+1\), \(dt=4x^3\,dx \Rightarrow x^3\,dx = \dfrac{dt}{4}\).
\(\displaystyle\frac{1}{4}\int t^{-3}\,dt = \frac{1}{4}\cdot\frac{t^{-2}}{-2}+C = -\frac{1}{8(x^4+1)^2}+C\)
Q4. Evaluate \(\displaystyle\int \tan x\,dx\)
Solution (Derivation of Formula 9):
Write \(\tan x = \dfrac{\sin x}{\cos x}\). Let \(t=\cos x\), \(dt = -\sin x\,dx\).
\(\displaystyle\int\frac{-dt}{t} = -\ln|t|+C = -\ln|\cos x|+C = \boxed{\ln|\sec x|+C}\)
Q5. Evaluate \(\displaystyle\int \frac{e^{\sqrt{x}}}{\sqrt{x}}\,dx\)
Solution:
Let \(t=\sqrt{x}\), \(dt=\dfrac{1}{2\sqrt{x}}dx \Rightarrow \dfrac{dx}{\sqrt{x}}=2\,dt\).
\(\displaystyle\int e^t \cdot 2\,dt = 2e^t+C = \boxed{2e^{\sqrt{x}}+C}\)
Q6. Evaluate \(\displaystyle\int \frac{dx}{\sqrt{1-9x^2}}\)
Solution:
Rewrite: \(\displaystyle\int\frac{dx}{\sqrt{1-(3x)^2}} = \frac{1}{3}\int\frac{3\,dx}{\sqrt{1-(3x)^2}} =
\frac{1}{3}\sin^{-1}(3x)+C\)
OR use formula 21 with \(a=\dfrac{1}{3}\): \(\sin^{-1}\!\dfrac{x}{1/3}+C =
\boxed{\dfrac{1}{3}\sin^{-1}(3x)+C}\)
Q7. Evaluate \(\displaystyle\int \frac{x\,dx}{x^4+16}\)
Solution:
Let \(t=x^2\), \(dt=2x\,dx\).
\(\displaystyle\frac{1}{2}\int\frac{dt}{t^2+16} = \frac{1}{2}\cdot\frac{1}{4}\tan^{-1}\!\frac{t}{4}+C =
\boxed{\frac{1}{8}\tan^{-1}\!\frac{x^2}{4}+C}\)
Q8. Evaluate \(\displaystyle\int \sin^5 x \cos x\,dx\)
Solution:
Pattern \([f(x)]^n \cdot f'(x)\) with \(f(x)=\sin x\), \(f'(x)=\cos x\), \(n=5\).
\(\displaystyle\int (\sin x)^5\cos x\,dx = \frac{\sin^6 x}{6}+C\)
Q9. Evaluate \(\displaystyle\int \frac{1+\ln x}{x(2+\ln x)}\,dx\)
Solution:
Let \(t=\ln x\), \(dt=\dfrac{dx}{x}\). Integral becomes \(\displaystyle\int\frac{1+t}{2+t}\,dt =
\int\left(1-\frac{1}{t+2}\right)dt = t-\ln|t+2|+C\)
\(= \boxed{\ln x - \ln|\ln x + 2|+C}\)
Q10. Evaluate \(\displaystyle\int \frac{\sin^{-1}x}{\sqrt{1-x^2}}\,dx\)
Solution:
Let \(t=\sin^{-1}x\), \(dt = \dfrac{dx}{\sqrt{1-x^2}}\). Integral \(= \displaystyle\int t\,dt =
\dfrac{t^2}{2}+C = \boxed{\dfrac{(\sin^{-1}x)^2}{2}+C}\)
Chapter 5: Special Substitutions for Algebraic Roots
\(\sqrt{a^2-x^2}\)
Use \(x = a\sin\theta\) or \(x =
a\cos\theta\).
Then \(\sqrt{a^2-x^2} = a\cos\theta\).
\(\sqrt{a^2+x^2}\)
Use \(x = a\tan\theta\).
Then \(\sqrt{a^2+x^2} =
a\sec\theta\).
\(\sqrt{x^2-a^2}\)
Use \(x = a\sec\theta\).
Then \(\sqrt{x^2-a^2} =
a\tan\theta\).
\(\sqrt{\dfrac{a-x}{a+x}}\)
Use \(x = a\cos 2\theta\).
Q11. Evaluate \(\displaystyle\int \frac{dx}{x\sqrt{x^2-1}}\)
Solution using formula 19 variant:
Let \(x=\sec\theta\), \(dx=\sec\theta\tan\theta\,d\theta\), \(\sqrt{x^2-1}=\tan\theta\).
\(\displaystyle\int\frac{\sec\theta\tan\theta\,d\theta}{\sec\theta\cdot\tan\theta} = \int d\theta = \theta+C
= \sec^{-1}x+C\)
Q12. Evaluate \(\displaystyle\int \sqrt{4-x^2}\,dx\)
Solution using Formula 25 with \(a=2\):
\(\displaystyle = \frac{x}{2}\sqrt{4-x^2}+\frac{4}{2}\sin^{-1}\!\frac{x}{2}+C =
\boxed{\frac{x}{2}\sqrt{4-x^2}+2\sin^{-1}\!\frac{x}{2}+C}\)
Q13. Evaluate \(\displaystyle\int \frac{dx}{\sqrt{x^2+4x+13}}\)
Solution — Complete the square first:
\(x^2+4x+13 = (x+2)^2+9\). Now use formula 24 with \(x \to (x+2)\), \(a=3\):
\(\displaystyle = \ln\!\left|(x+2)+\sqrt{(x+2)^2+9}\right|+C =
\boxed{\ln\!\left|x+2+\sqrt{x^2+4x+13}\right|+C}\)
Q14. Evaluate \(\displaystyle\int \frac{dx}{\sqrt{3-4x-x^2}}\)
Solution — Complete the square:
\(3-4x-x^2 = -(x^2+4x-3) = -[(x+2)^2-7] = 7-(x+2)^2\)
Use formula 21 with \(u=x+2\), \(a=\sqrt7\):
\(\displaystyle = \sin^{-1}\!\frac{x+2}{\sqrt{7}}+C\)
Q15. Evaluate \(\displaystyle\int \frac{x+1}{\sqrt{x^2+2x+5}}\,dx\)
Solution — Numerator = \(\frac{1}{2}\) × derivative of \(x^2+2x+5\):
\(x+1 = \frac{1}{2}(2x+2)\). So the integral is
\(\dfrac{1}{2}\displaystyle\int\frac{2x+2}{\sqrt{x^2+2x+5}}\,dx = \sqrt{x^2+2x+5}+C\)
Chapter 6: Key Error Traps
Error Trap 1
\(\int\frac{1}{x^2}\,dx \ne \ln x^2 + C\)
\(\frac{1}{x^2} = x^{-2}\). Correct result: \(\displaystyle\int x^{-2}\,dx = \frac{x^{-1}}{-1}+C =
-\frac{1}{x}+C\)
Error Trap 2
\(\int\frac{f'(x)}{f(x)}dx = \ln|f(x)|+C\) — the modulus is necessary!
Writing \(\ln f(x)\) without modulus is incorrect when \(f(x)\) can be negative.
Error Trap 3
Do not confuse \(\int\sin^2 x\,dx\) with \(\frac{\sin^3x}{3}+C\)
\(\sin^2 x\) is NOT in the "power of sin" chain rule form because the outer derivative of \(\sin x\) is
\(\cos x\), absent here. Correct method: use \(\sin^2 x = \frac{1-\cos 2x}{2}\).
Vardaan Learning Institute · IC Module 1/5 · JEE Masterclass 2026–27
