MathIsimple
Course 9

Antiderivatives and Basic Integration

Section 1: Antiderivatives

Definition 1.1: Antiderivative

Let f(x)f(x) be a function defined on an interval II. A function F(x)F(x) is called an antiderivative of f(x)f(x) on II if:

F′(x)=f(x)for all x∈IF'(x) = f(x) \quad \text{for all } x \in I
Theorem 1.1: Antiderivatives Differ by Constant

If F(x)F(x) and G(x)G(x) are both antiderivatives of f(x)f(x) on II, then:

F(x)−G(x)=CF(x) - G(x) = C

for some constant CC.

Proof of Theorem 1.1:

Since F′(x)=G′(x)=f(x)F'(x) = G'(x) = f(x), we have:

(F−G)′(x)=F′(x)−G′(x)=0(F - G)'(x) = F'(x) - G'(x) = 0

By MVT, F(x)−G(x)F(x) - G(x) is constant. ∎

∎
Definition 1.2: Indefinite Integral

The indefinite integral of f(x)f(x) is:

∫f(x) dx=F(x)+C\int f(x)\,dx = F(x) + C

where F(x)F(x) is any antiderivative of f(x)f(x) and CC is the constant of integration.

Section 2: Basic Integration Formulas

Power Rule

For n≠−1n \neq -1:

∫xn dx=xn+1n+1+C\int x^n\,dx = \frac{x^{n+1}}{n+1} + C

For n=−1n = -1:

∫1x dx=ln⁡∣x∣+C\int \frac{1}{x}\,dx = \ln|x| + C
Exponential Functions
∫ex dx=ex+C\int e^x\,dx = e^x + C
∫ax dx=axln⁡a+C(a>0,a≠1)\int a^x\,dx = \frac{a^x}{\ln a} + C \quad (a > 0, a \neq 1)
Trigonometric Functions
∫sin⁡x dx=−cos⁡x+C\int \sin x\,dx = -\cos x + C
∫cos⁡x dx=sin⁡x+C\int \cos x\,dx = \sin x + C
∫sec⁡2x dx=tan⁡x+C\int \sec^2 x\,dx = \tan x + C
∫csc⁡2x dx=−cot⁡x+C\int \csc^2 x\,dx = -\cot x + C
∫sec⁡xtan⁡x dx=sec⁡x+C\int \sec x \tan x\,dx = \sec x + C
∫csc⁡xcot⁡x dx=−csc⁡x+C\int \csc x \cot x\,dx = -\csc x + C
Inverse Trigonometric Functions
∫11−x2 dx=arcsin⁡x+C\int \frac{1}{\sqrt{1-x^2}}\,dx = \arcsin x + C
∫11+x2 dx=arctan⁡x+C\int \frac{1}{1+x^2}\,dx = \arctan x + C
Example 2.1: Basic Integration Examples

(a) ∫x7 dx=x88+C\int x^7\,dx = \frac{x^8}{8} + C

(b) ∫e3x dx=e3x3+C\int e^{3x}\,dx = \frac{e^{3x}}{3} + C

(c) ∫cos⁡x dx=sin⁡x+C\int \cos x\,dx = \sin x + C

(d) ∫11+x2 dx=arctan⁡x+C\int \frac{1}{1+x^2}\,dx = \arctan x + C

Section 3: Linearity Properties

Theorem 3.1: Linearity of Integration
∫[f(x)+g(x)] dx=∫f(x) dx+∫g(x) dx\int [f(x) + g(x)]\,dx = \int f(x)\,dx + \int g(x)\,dx
∫kf(x) dx=k∫f(x) dx\int kf(x)\,dx = k\int f(x)\,dx
Example 3.1: Linearity Application

Find: ∫(3x2+2sin⁡x−ex) dx\int (3x^2 + 2\sin x - e^x)\,dx

∫(3x2+2sin⁡x−ex) dx=3∫x2 dx+2∫sin⁡x dx−∫ex dx\int (3x^2 + 2\sin x - e^x)\,dx = 3\int x^2\,dx + 2\int \sin x\,dx - \int e^x\,dx
=x3−2cos⁡x−ex+C= x^3 - 2\cos x - e^x + C

Section 4: Trigonometric Integrals

Example 4.1: Trigonometric Integrals
  • ∫sin⁡x dx=−cos⁡x+C\int \sin x\,dx = -\cos x + C
  • ∫cos⁡x dx=sin⁡x+C\int \cos x\,dx = \sin x + C
  • ∫sec⁡2x dx=tan⁡x+C\int \sec^2 x\,dx = \tan x + C
  • ∫csc⁡2x dx=−cot⁡x+C\int \csc^2 x\,dx = -\cot x + C

Section 5: Exponential and Logarithmic Integrals

Example 5.1: Exponential Integrals
  • ∫ex dx=ex+C\int e^x\,dx = e^x + C
  • ∫ax dx=axln⁡a+C\int a^x\,dx = \frac{a^x}{\ln a} + C
  • ∫1x dx=ln⁡∣x∣+C\int \frac{1}{x}\,dx = \ln |x| + C

Section 6: Inverse Trigonometric Integrals

Example 6.1: Inverse Trig Integrals
  • ∫11−x2 dx=arcsin⁡x+C\int \frac{1}{\sqrt{1-x^2}}\,dx = \arcsin x + C
  • ∫11+x2 dx=arctan⁡x+C\int \frac{1}{1+x^2}\,dx = \arctan x + C
  • ∫1∣x∣x2−1 dx=arcsec x+C\int \frac{1}{|x|\sqrt{x^2-1}}\,dx = \text{arcsec } x + C

Section 7: Integration by Recognition

Example 7.1: Recognition Method

Find: ∫2x1+x2 dx\int \frac{2x}{1+x^2}\,dx

Solution: Recognize that (1+x2)′=2x(1+x^2)' = 2x:

∫2x1+x2 dx=ln⁡(1+x2)+C\int \frac{2x}{1+x^2}\,dx = \ln(1+x^2) + C

Work These Ideas by Hand — Then Check Yourself

Practice Quiz: Antiderivatives and Basic Integration
10
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1
If F′(x)=f(x)F'(x) = f(x) for all x∈Ix \in I, then F(x)F(x) is called:
Easy
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2
If F(x)F(x) and G(x)G(x) are both antiderivatives of f(x)f(x), then:
Easy
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3
∫x5 dx=\int x^5\,dx =
Easy
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4
∫e3x dx=\int e^{3x}\,dx =
Easy
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5
∫tan⁡x dx=\int \tan x\,dx =
Medium
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6
∫11+x2 dx=\int \frac{1}{1+x^2}\,dx =
Easy
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7
∫3x dx=\int 3^x\,dx =
Medium
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8
∫11−x2 dx=\int \frac{1}{\sqrt{1-x^2}}\,dx =
Medium
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9
∫[f(x)+g(x)] dx=\int [f(x) + g(x)]\,dx =
Medium
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10
∫0 dx=\int 0\,dx =
Easy
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Frequently Asked Questions

What is the difference between antiderivative and indefinite integral?

An antiderivative F(x) is a specific function with F'(x) = f(x). The indefinite integral ∫f(x)dx is the family of all antiderivatives: F(x) + C.

Why do we add '+C' to indefinite integrals?

Because if F(x) is an antiderivative, so is F(x) + C for any constant. The '+C' represents all possible antiderivatives.

Does every function have an antiderivative?

No. But continuous functions always have antiderivatives (Fundamental Theorem). Functions with jump discontinuities may not.

How do I know which formula to use?

Identify the form of the integrand. Look for: power functions (x^n), exponentials (e^x, a^x), trig functions, inverse trig patterns (1/√(1-x²), 1/(1+x²)), and logarithms (1/x).

What is the power rule for integration?

For n ≠ -1: ∫x^n dx = x^(n+1)/(n+1) + C. For n = -1: ∫1/x dx = ln|x| + C.