Quick answer: An indefinite integral ∫ f ( x ) d x = F ( x ) + C \int f(x)\,dx = F(x) + C ∫ f ( x ) d x = F ( x ) + C F ′ ( x ) = f ( x ) F'(x) = f(x) F ′ ( x ) = f ( x ) ∫ x n d x = x n + 1 n + 1 + C \int x^n\,dx = \frac{x^{n + 1}}{n + 1} + C ∫ x n d x = n + 1 x n + 1 + C n ≠ − 1 n \ne -1 n = − 1 ∫ 1 x d x = ln ∣ x ∣ + C \int\frac{1}{x}\,dx = \ln\lvert x\rvert + C ∫ x 1 d x = ln ∣ x ∣ + C ∫ e x d x = e x + C \int e^x\,dx = e^x + C ∫ e x d x = e x + C ∫ sin x d x = − cos x + C \int\sin x\,dx = -\cos x + C ∫ sin x d x = − cos x + C substitution , put t = g ( x ) t = g(x) t = g ( x ) d t = g ′ ( x ) d x dt = g'(x)\,dx d t = g ′ ( x ) d x ∫ f ′ ( x ) f ( x ) d x = ln ∣ f ( x ) ∣ + C \int\frac{f'(x)}{f(x)}\,dx = \ln\lvert f(x)\rvert + C ∫ f ( x ) f ′ ( x ) d x = ln ∣ f ( x )∣ + C ∫ [ f ( x ) ] n f ′ ( x ) d x = [ f ( x ) ] n + 1 n + 1 + C \int[f(x)]^nf'(x)\,dx = \frac{[f(x)]^{n + 1}}{n + 1} + C ∫ [ f ( x ) ] n f ′ ( x ) d x = n + 1 [ f ( x ) ] n + 1 + C
Integration reverses the standard derivatives .
Standard integrals
∫ f ( x ) d x \int f(x)\,dx ∫ f ( x ) d x Result (+ C) x n x^n x n n ≠ − 1 n \ne -1 n = − 1 x n + 1 n + 1 \frac{x^{n + 1}}{n + 1} n + 1 x n + 1 1 x \frac{1}{x} x 1 ln ∣ x ∣ \ln\lvert x\rvert ln ∣ x ∣ e x e^x e x e x e^x e x a x a^x a x a x ln a \frac{a^x}{\ln a} l n a a x sin x \sin x sin x − cos x -\cos x − cos x cos x \cos x cos x sin x \sin x sin x sec 2 x \sec^2 x sec 2 x tan x \tan x tan x csc 2 x \csc^2 x csc 2 x − cot x -\cot x − cot x sec x tan x \sec x\tan x sec x tan x sec x \sec x sec x csc x cot x \csc x\cot x csc x cot x − csc x -\csc x − csc x tan x \tan x tan x ln ∣ sec x ∣ \ln\lvert\sec x\rvert ln ∣ sec x ∣ cot x \cot x cot x ln ∣ sin x ∣ \ln\lvert\sin x\rvert ln ∣ sin x ∣ sec x \sec x sec x ln ∣ sec x + tan x ∣ \ln\lvert\sec x + \tan x\rvert ln ∣ sec x + tan x ∣ csc x \csc x csc x ln ∣ csc x − cot x ∣ \ln\lvert\csc x - \cot x\rvert ln ∣ csc x − cot x ∣ 1 1 − x 2 \frac{1}{\sqrt{1 - x^2}} 1 − x 2 1 sin − 1 x \sin^{-1}x sin − 1 x 1 1 + x 2 \frac{1}{1 + x^2} 1 + x 2 1 tan − 1 x \tan^{-1}x tan − 1 x
Properties
Property Statement Linearity ∫ [ a f ( x ) + b g ( x ) ] d x = a ∫ f d x + b ∫ g d x \int[af(x) + bg(x)]\,dx = a\int f\,dx + b\int g\,dx ∫ [ a f ( x ) + b g ( x )] d x = a ∫ f d x + b ∫ g d x Linear argument If ∫ f ( x ) d x = F ( x ) \int f(x)\,dx = F(x) ∫ f ( x ) d x = F ( x ) ∫ f ( a x + b ) d x = 1 a F ( a x + b ) \int f(ax + b)\,dx = \frac{1}{a}F(ax + b) ∫ f ( a x + b ) d x = a 1 F ( a x + b )
Integration by substitution
Method:
Step 1: Choose t = g ( x ) t = g(x) t = g ( x )
Step 2: Replace g ′ ( x ) d x g'(x)\,dx g ′ ( x ) d x d t dt d t
Step 3: Integrate in t, then substitute back.
Form Result ∫ f ′ ( x ) f ( x ) d x \int\frac{f'(x)}{f(x)}\,dx ∫ f ( x ) f ′ ( x ) d x ln ∣ f ( x ) ∣ + C \ln\lvert f(x)\rvert + C ln ∣ f ( x )∣ + C ∫ [ f ( x ) ] n f ′ ( x ) d x \int[f(x)]^nf'(x)\,dx ∫ [ f ( x ) ] n f ′ ( x ) d x [ f ( x ) ] n + 1 n + 1 + C \frac{[f(x)]^{n + 1}}{n + 1} + C n + 1 [ f ( x ) ] n + 1 + C ∫ e f ( x ) f ′ ( x ) d x \int e^{f(x)}f'(x)\,dx ∫ e f ( x ) f ′ ( x ) d x e f ( x ) + C e^{f(x)} + C e f ( x ) + C ∫ f ′ ( x ) f ( x ) d x \int\frac{f'(x)}{\sqrt{f(x)}}\,dx ∫ f ( x ) f ′ ( x ) d x 2 f ( x ) + C 2\sqrt{f(x)} + C 2 f ( x ) + C
Trigonometric identities in integration
Integrand Rewrite as sin 2 x \sin^2 x sin 2 x 1 − cos 2 x 2 \frac{1 - \cos 2x}{2} 2 1 − c o s 2 x cos 2 x \cos^2 x cos 2 x 1 + cos 2 x 2 \frac{1 + \cos 2x}{2} 2 1 + c o s 2 x tan 2 x \tan^2 x tan 2 x sec 2 x − 1 \sec^2 x - 1 sec 2 x − 1 sin m x cos n x \sin mx\cos nx sin m x cos n x Product-to-sum formula sin 3 x \sin^3 x sin 3 x 3 sin x − sin 3 x 4 \frac{3\sin x - \sin 3x}{4} 4 3 s i n x − s i n 3 x
The product-to-sum step uses the formulas in product-to-sum and sum-to-product .
Worked examples
Example 1: power rule
∫ ( 3 x 2 + 2 x − e x ) d x = x 3 + 2 ln ∣ x ∣ − e x + C \int\left(3x^2 + \frac{2}{x} - e^x\right)dx = x^3 + 2\ln\lvert x\rvert - e^x + C ∫ ( 3 x 2 + x 2 − e x ) d x = x 3 + 2 ln ∣ x ∣ − e x + C
∫ 2 x + 3 x 2 + 3 x + 5 d x = ln ∣ x 2 + 3 x + 5 ∣ + C \int\frac{2x + 3}{x^2 + 3x + 5}\,dx = \ln\lvert x^2 + 3x + 5\rvert + C ∫ x 2 + 3 x + 5 2 x + 3 d x = ln ∣ x 2 + 3 x + 5 ∣ + C
Example 3: substitution
Evaluate ∫ x e x 2 d x \int x\,e^{x^2}\,dx ∫ x e x 2 d x
Put t = x 2 t = x^2 t = x 2 d t = 2 x d x dt = 2x\,dx d t = 2 x d x 1 2 ∫ e t d t = 1 2 e x 2 + C \frac{1}{2}\int e^t\,dt = \frac{1}{2}e^{x^2} + C 2 1 ∫ e t d t = 2 1 e x 2 + C
Example 4: power of a function
Evaluate ∫ sin 3 x cos x d x \int\sin^3 x\cos x\,dx ∫ sin 3 x cos x d x
Put t = sin x t = \sin x t = sin x ∫ t 3 d t = sin 4 x 4 + C \int t^3\,dt = \frac{\sin^4 x}{4} + C ∫ t 3 d t = 4 s i n 4 x + C
Example 5: trig identity
∫ sin 2 x d x = ∫ 1 − cos 2 x 2 d x = x 2 − sin 2 x 4 + C \int\sin^2 x\,dx = \int\frac{1 - \cos 2x}{2}\,dx = \frac{x}{2} - \frac{\sin 2x}{4} + C ∫ sin 2 x d x = ∫ 2 1 − c o s 2 x d x = 2 x − 4 s i n 2 x + C
Example 6: linear argument
∫ cos ( 3 x + 2 ) d x = 1 3 sin ( 3 x + 2 ) + C \int\cos(3x + 2)\,dx = \frac{1}{3}\sin(3x + 2) + C ∫ cos ( 3 x + 2 ) d x = 3 1 sin ( 3 x + 2 ) + C
Summary table
Idea Key result Power rule x n + 1 n + 1 \frac{x^{n + 1}}{n + 1} n + 1 x n + 1 1 x \frac{1}{x} x 1 ln ∣ x ∣ \ln\lvert x\rvert ln ∣ x ∣ f ′ f \frac{f'}{f} f f ′ ln ∣ f ∣ \ln\lvert f\rvert ln ∣ f ∣ f ( a x + b ) f(ax + b) f ( a x + b ) Divide by a sin 2 x \sin^2 x sin 2 x Use 1 − cos 2 x 2 \frac{1 - \cos 2x}{2} 2 1 − c o s 2 x
Exam tips
Look for a function whose derivative also appears; that is your substitution.
Check answers by differentiating.
Do not forget the constant of integration.
Next, learn integration by parts and partial fractions , or see the NIMCET Calculus chapter .
Common mistakes
Applying the power rule to 1 x \frac{1}{x} x 1
Forgetting to divide by a for f ( a x + b ) f(ax + b) f ( a x + b )
Leaving the answer in terms of t after substitution.
Dropping the absolute value in ln ∣ x ∣ \ln\lvert x\rvert ln ∣ x ∣
Frequently Asked Questions
What is the integral of 1/x?
ln|x| + C.
What is the integral of f′(x)/f(x)?
ln|f(x)| + C.
How do you choose a substitution?
Pick an inner function whose derivative also appears in the integrand, up to a constant factor.
What is the integral of sin²x?
x/2 − sin 2x/4 + C.
What is the integral of sec x?
ln|sec x + tan x| + C.
