In the previous discussions, we learned how to find the derivatives of different functions.

Now, we will
introduce the inverse of differentiation. We shall call this process antidifferentiation. A natural question then
arises:

A function F is an antiderivative of the function f on an interval I if :

F0(x) = f(x) for every value of x in I.

As previously discussed, the process of antidifferentiation is just the inverse process of finding the
derivatives of functions. We have shown in the previous lesson that a function can have a family of
antiderivatives.

We will look at antiderivatives of different types of functions. Particularly, we will find the
antiderivatives of polynomial functions, rational functions and radical functions.

Terminologies and Notations:

• Antidifferentation is the process of finding the antiderivative.
• The symbol ∫ , also called the integral sign, denotes the operation of antidifferentiation.
• The function f is called the integrand.
• If F is an antiderivative of f, we write∫ (𝑥)𝑑𝑥 = 𝐹(𝑥) + 𝐶.
• The symbols ∫ and dx go hand-in-hand and dx helps us identify the variable of integration.
• The expression F(x)+C is called the general antiderivative of f. Meanwhile, each antiderivative of f is
called a particular antiderivative of f.
DEVELOPMENT
We will now give examples of antiderivatives of functions.

EXAMPLE 1:
(a) An antiderivative of f(x) = 12x2 + 2x is F(x)=4x3 + x2 .
As we can see, the derivative of F is given by F0(x) = 12x2 + 2x = f(x).

(b) An antiderivative of g(x) = cos x is G(x) = sin x because G0(x) = cos x = g(x).
Remark 1: The antiderivative F of a function f is not unique.

EXAMPLE 2:
(a) Other antiderivatives of f(x) = 12x2 + 2x are F1(x)=4x3 + x2 – 1 and F2(x) = 4x3 + x2 + 1.
In fact, any function of the form F(x)=4x3 + x2 + C, where C ∈ R is an antiderivative of f(x).
Observe that F0(x) = 12x2 + 2x + 0 = 12x2 + 2x = f(x).

(b) Other antiderivatives of g(x) = cos x are G1(x) = sin x + 𝜋 and G2(x) = sin x – 1.
In fact, any function G(x) = sin x + C, where C∈ R is an antiderivative of g(x).

Theorem 10.
If F is an antiderivative of f on an interval I, then every antiderivative of f on I is given by F(x) + C, where
C is an arbitrary constant.

Remark 2: Using the theorem above, we can conclude that if F1 and F2 are antiderivatives of f, then F2 (x)
= F1 (x) + C. That is, F1 and F2 differ only by a constant.
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Let us recall first the following differentiation formulas:
(a) Dx(x)=1
(b) Dx(xn) = nxn – 1, where n is any real number
(c) (c) Dx[a(f(x))] = aDx[f(x)]
(d) Dx[f(x) ± g(x)] = Dx[f(x)] ± Dx[g(x)]
The above formulas lead to the following theorem which are used in obtaining the antiderivatives of
functions. We apply them to integrate polynomials, rational functions and radical functions.

Theorem 11. (Theorems on Antidifferentiation)

Antiderivatives are the opposite of derivatives. An antiderivative is a function that reverses what the
derivative does. One function has many antiderivatives, but they all take the form of a function plus an
arbitrary constant. Antiderivatives are a key part of indefinite integrals.
LEARNING TASK: (1 yellow paper)
Compute for the Indefinite Integrals of the following. Show your complete solution if needed.
1. ∫ 827𝑑𝑥 2. ∫ 𝑥 2 𝑑𝑥
3. ∫(𝑥 3 + 2𝑥 2 )𝑑𝑥 4. ∫(3𝑥 2 + 1)𝑑𝑥
5 4
5. ∫(3𝑥 + 4𝑥 + 6𝑥 + 3)𝑑𝑥 6. ∫(𝑥 −4 )𝑑𝑥

INTEGRATION THEOREMS
∫ 0𝑑𝑥 = 𝐶 ∫ cos 𝑥 𝑑𝑥 = sin 𝑥 + 𝐶

∫ 𝑘𝑑𝑥 = 𝑘𝑥 + 𝐶 ∫ sin 𝑥 𝑑𝑥 = −cos 𝑥 + 𝐶

∫ 𝑘𝑓(𝑥)𝑑𝑥 = 𝑘 ∫ 𝑓(𝑥)𝑑𝑥 ∫ 𝑠𝑒𝑐 2 𝑥 𝑑𝑥 = tan 𝑥 + 𝐶

∫[𝑓(𝑥) ± 𝑔(𝑥)]𝑑𝑥 = ∫ 𝑓(𝑥)𝑑𝑥 ± ∫ 𝑔(𝑥)𝑑𝑥 ∫ 𝑐𝑠𝑐 2 𝑥 𝑑𝑥 = −cot 𝑥 + 𝐶

𝑥 𝑛+1
∫ 𝑥 𝑛 𝑑𝑥 = +𝐶 (𝑛 ≠ −1) ∫ sec 𝑥 tan 𝑥 𝑑𝑥 = sec 𝑥 + 𝐶
𝑛+1
1
∫ 𝑑𝑥 = ln 𝑥 + 𝐶 ∫ 𝑐𝑠𝑐 𝑥 cot 𝑥 𝑑𝑥 = −csc 𝑥 + 𝐶
𝑥
𝑥 𝑥 𝑥
𝑏𝑥
∫ 𝑒 𝑑𝑥 = 𝑒 + 𝐶 ∫ 𝑏 𝑑𝑥 = +𝐶 (𝑏 > 0, 𝑏 ≠ 1)
ln 𝑏

Page 2 of 4
Examples:
1. ∫ 5283 𝑑𝑥 = 5283𝑥 + 𝐶

2. Steps in getting antiderivatives

∫ 7𝑥𝑑𝑥 = 7 ∫ 𝑥𝑑𝑥 Factor out the constant
= 7 ∫ 𝑥 1 𝑑𝑥 x can be written as 𝑥 1
𝑥 1+1
= 7 ( 1+1 ) Apply the Integration Theorem (think of the opposite procedure when you get the derivative)
𝑥2
= 7(2 ) Simplify
7𝑥 2
= +𝐶 Add C which represents the arbitrary constant
2

3. Steps in getting the antiderivative of simple integrals

1
a. ∫ 𝑥 5 𝑑𝑥 = ∫ 𝑥 −5 𝑑𝑥 Rewrite using negative exponent
𝑥 −5+1
= +𝐶 Apply the Integration Theorem
−5+1
*You can already Add C which represents the arbitrary constant
𝑥 −4
= +𝐶 Simplify (Add)
−4
1
= − 4𝑥 4 +𝐶 Simplify (Apply the definition of negative exponent)

2
5
𝑏. ∫ √𝑥 2 𝑑𝑥 = ∫ 𝑥 5 𝑑𝑥 Rewrite using rational exponent
2
+1
𝑥5
= 2 +𝐶 Apply the Integration Theorem
+1
5
*You can already Add C which represents the arbitrary constant
7
𝑥5
= 7 +𝐶 Simplify (Add)
5
7
5
= 7 𝑥5 + 𝐶 Simplify

c. ∫ 3𝑠𝑒𝑐 2 𝑥𝑑𝑥 = 3 ∫ 𝑠𝑒𝑐 2 𝑥𝑑𝑥 Factor out the constant

= 3(tan 𝑥) + 𝐶 Apply the Integration Theorem
*You can already Add C which represents the arbitrary constant
= 3𝑡𝑎𝑛𝑥 + 𝐶 Simplify

4. Antidifferentiation of Polynomials
𝑎. ∫(𝑥 − 5)𝑑𝑥 = ∫ 𝑥𝑑𝑥 − ∫ 5𝑑𝑥 Get the antiderivative of each term
𝑥2
= + 𝐶1 − 5𝑥 + 𝐶2 Note: C=𝐶1 + 𝐶2
2
𝑥2
= − 5𝑥 + 𝐶
2

𝑏. ∫(5𝑥 4 + 2𝑥 2 − 3𝑥)𝑑𝑥
= ∫ 5𝑥 4 𝑑𝑥 + ∫ 2𝑥 2 𝑑𝑥 − ∫ 3𝑥𝑑𝑥 Get the antiderivative of each term
= 5 ∫ 𝑥 4 𝑑𝑥 + 2 ∫ 𝑥 2 𝑑𝑥 − 3 ∫ 𝑥𝑑𝑥 Factor out the constant
5𝑥 5 2𝑥 3 3𝑥 2
= + 3 − 2 +𝐶 Apply the Integration Theorem
5
2𝑥 3 3𝑥 2
= 𝑥5 + 3 − 2 + 𝐶 Simplify

Page 3 of 4
 6𝑥 5 + 4𝑥 3 − 9𝑥 2
= +𝐶
6
2 (6𝑥 3
𝑥 + 4𝑥 − 9)
= +𝐶
6

𝑐. ∫(−14𝑥 + 6)𝑑𝑥
= ∫ −14𝑥𝑑𝑥 + ∫ 6𝑑𝑥 Get the antiderivative of each term
= −14 ∫ 𝑥𝑑𝑥 + 6 ∫ 𝑑𝑥 Factor out the constant
−14𝑥 1+1
= + 6𝑥 + 𝐶 Apply the Integration Theorem
1+1
−14𝑥 2
= + 6𝑥 + 𝐶 Simplify
2
2
= −7𝑥 + 6𝑥 + 𝐶

5. Solve for the indefinite integral of the following

𝑥+2 𝑥 2
a. ∫ ( 𝑥 ) 𝑑𝑥 = ∫ ( 𝑥 + 𝑥) 𝑑𝑥 Write as 2 separate fractions
√ √ √
1 1
1−2
= ∫ (𝑥 + 2𝑥 −2 ) 𝑑𝑥 Apply laws of rational exponents

1 1
= ∫ (𝑥 2 + 2𝑥 −2 ) 𝑑𝑥 Simplify
3 1
𝑥2 2𝑥 2
= 3 + 1 +𝐶 Apply the Integration Theorem
2 2
3
1
2𝑥 2
= + 4𝑥 2 + 𝐶 Simplify
3
2𝑥√𝑥
= + 4√𝑥 + 𝐶 Change terms with rational exponents into radicals
3
𝑥
= 2√𝑥 (3 + 2) + 𝐶 Factor out the greatest common factor
𝑥+6
= 2√𝑥 ( 3 ) + 𝐶 Simplify

sin 𝑥 1 sin 𝑥 sin 𝑥

b. ∫ 𝑐𝑜𝑠2 𝑥 𝑑𝑥 = ∫ (cos 𝑥) (cos 𝑥) 𝑑𝑥 *𝑐𝑜𝑠 2 𝑥 is not found among the theorem that is why we rewrite them as 2 trig

= ∫ sec 𝑥 tan 𝑥 𝑑𝑥 Apply reciprocal identity for secant & quotient identity for tangent
= sec 𝑥 + 𝐶 Apply the Integration Theorem

c. ∫(𝑥 2 − 1)2 𝑑𝑥 = ∫(𝑥 4 − 2𝑥 2 + 1)𝑑𝑥 Expand

𝑥5 2𝑥 3
= − +𝑥+𝐶 Apply the Integration Theorem
5 3

3 5𝑥
d. ∫ (2𝑒 𝑥 − 𝑥 + ) 𝑑𝑥
2
3 5𝑥
= ∫ 2𝑒 𝑥 𝑑𝑥 − ∫ 𝑥 𝑑𝑥 + ∫ 2 𝑑𝑥 Get the integral of each term
1 1
= 2 ∫ 𝑒 𝑥 𝑑𝑥 − 3 ∫ 𝑥 𝑑𝑥 + 2 ∫ 5𝑥 𝑑𝑥 Factor out the constant
5𝑥
= 2𝑒 𝑥 − 3 ln 𝑥 + 2 ln 5 + 𝐶 Apply the Integration Theorem

Page 4 of 4