The document defines integration as the inverse operation of differentiation or the antiderivative. Integration finds the function given its derivative, while differentiation finds the derivative of a function. The key points are: 1) Integration is denoted by the integral sign ∫ and finds the antiderivative F(x) of a function f(x) plus a constant c. 2) Some basic integration rules and theorems are presented, including formulas for integrating polynomials and trigonometric functions. 3) The substitution rule is described for performing integral substitutions to solve integrals that can't be solved with basic formulas. Examples of integrating trigonometric functions and expressions involving square roots are provided.

1. A. Definition of Integration
In Class XI, you have learned the concept of derivative. Comprehension
on the derivative concept you can use to understand
integration concept. For that, try to determine the following derivative functions:
 𝑓1(𝑥) = 3𝑥3
+ 3
 𝑓2(𝑥) = 3𝑥3
+ 7
 𝑓3(𝑥) = 3𝑥3
− 1
 𝑓4(𝑥) = 3𝑥3
− 10
 𝑓5(𝑥) = 3𝑥3
− 99
Note that these functions have the general form , 𝑓(𝑥) = 3𝑥3
+ 𝑐 with c is constant. Each
function has a derivative 𝑓′(𝑥) = 9𝑥2
. Thus, the derivative function 𝑓(𝑥) = 3𝑥3
+ 𝑐 is 𝑓′(𝑥) =
9𝑥2
.
Now,what if you have to define the function 𝑓(𝑥) of
𝑓′(𝑥) is known?Determine the function 𝑓(𝑥) from 𝑓′(𝑥), means determining
antiderivative of 𝑓′(𝑥). Thus, the integration is the antiderivative
(Antidiferensial) or the inverse operation of the differential.
If 𝐹(𝑥) is general function y that is common 𝐹′
(𝑥) = 𝑓(𝑥), then 𝐹(𝑥) is antiderivative
or integral of 𝑓(𝑥).
Integration function 𝑓(𝑥) with respect to 𝑥 is denoted as follows:
∫ 𝑓(𝑥) 𝑑𝑥 = 𝐹(𝑥) + 𝑐
With :
∫ = integration
𝑓(𝑥) = function integration
𝐹(𝑥) = integration common function
𝑐 = constanta
Now, consider the derivative of the following functions ;
𝑔1(𝑥) = 𝑥, be obtained 𝑔1′(𝑥) = 1
So, if 𝑔1′(𝑥) = 1,then 𝑔1(𝑥) = ∫ 𝑔1
′ (𝑥) 𝑑𝑥 = 𝑥 + 𝑐
𝑔2(𝑥) =
1
2
𝑥2
, be obtained 𝑔2′(𝑥) = 𝑥
So,if 𝑔2′(𝑥) = 𝑥, then 𝑔2(𝑥) = ∫ 𝑔2
′ (𝑥) 𝑑𝑥 =
1
2
𝑥2
+ 𝑐
𝑔3(𝑥) =
1
3
𝑥3
, be obtained 𝑔3′(𝑥) = 𝑥
So,if 𝑔3′(𝑥) = 𝑥, then 𝑔3(𝑥) = ∫ 𝑔3 ′(𝑥) 𝑑𝑥 =
1
3
𝑥3
+ 𝑐

2. 𝑔4(𝑥) =
1
6
𝑥6
,be obtained 𝑔4′(𝑥) = 𝑥5
So,if 𝑔4′(𝑥) = 𝑥5
, then 𝑔4(𝑥) = ∫ 𝑔4 ′(𝑥) 𝑑𝑥 =
1
6
𝑥6
+ 𝑐
Of this description, it appears that if 𝑔′(𝑥) = 𝑥 𝑛
, then 𝑔(𝑥) =
1
𝑛+1
𝑥 𝑛+1
+ 𝑐 or
can be written ∫ 𝑥 𝑛
𝑑𝑥 =
1
𝑛+1
𝑥 𝑛+1
+ 𝑐, 𝑛 ≠ −1.
For example, the derivative function 𝑓(𝑥) = 3𝑥3
+ 𝑐 is 𝑓′(𝑥) = 9𝑥2
.
This means, antiderivative of 𝑓′(𝑥) = 9𝑥2
is 𝑓(𝑥) = 3𝑥3
+ 𝑐 or written ∫ 𝑓′(𝑥)𝑑𝑥 = 3𝑥2
+ 𝑐.
This description illustrates the following relationship.
If 𝑓′(𝑥) = 𝑥 𝑛
,then 𝑓(𝑥) =
1
𝑛+1
𝑥 𝑛+1
+ 𝑐 , 𝑛 ≠ −1, with c is a constant.
Example:
1. Find the derivative of each of the following functions :
Answers:
2. Find the antiderivative x if known:
Answers:

3. B. Indefinite Integrals
In the previous part, you have known that the integral is an antiderivative. So, if there
is a function F(x) that can differential at intervals [𝑎, 𝑏] so that
𝑑(𝐹(𝑥))
𝑑𝑥
= 𝑓(𝑥),the
antiderivative of f (x) is F (x) + c.
Mathematically, written
∫ 𝑓(𝑥)𝑑𝑥 = 𝐹(𝑥) + 𝑐
where,∫ 𝑑𝑥 = symbol of stated integral antiderivative operation
f(x) = integrand functions, namely functions which sought antiderivative
c = constant
For example, you can write
Because,
So you can look at indefinite integral as representatives of the whole family of functions (one
antiderivative for each value constant c. The definition can be used to prove
the following theorems which will help in the execution of arithmetic
integrals.
Theorem 1
If n is a rational number and n ≠ −1,then ∫ 𝑥 𝑛
𝑑𝑥 =
1
𝑛+1
𝑥 𝑛+1
+ 𝑐 where
c is a constant.
Theorem 2
If f the integral function and k is a constant, then ∫ k f(x) dx = k ∫ f (x) dx
Theorem 3
If f and g is integral functions, then ∫ (𝑓(𝑥) + 𝑔(𝑥))𝑑𝑥 = ∫ 𝑓(𝑥)𝑑𝑥 + ∫ 𝑔(𝑥)𝑑𝑥
Theorem 4
If f and g is integral functions, then ∫ (𝑓(𝑥) − 𝑔(𝑥))𝑑𝑥 = ∫ 𝑓(𝑥)𝑑𝑥 − ∫ 𝑔(𝑥)𝑑𝑥
Theorem 5
Substitution Integrals Rule
If u is a function which can differential and r is a numbers which no zero, then
∫ (𝑢(𝑥))
𝑟
𝑢′(𝑥)𝑑𝑥 =
1
𝑛+1
(𝑢(𝑥))
𝑟+1
+ 𝑐, where c is a constant and r≠ −1

4. Theorem 6
Partial Integrals Rule
If u and v is a functions which can differential, then∫ 𝑢 𝑑𝑣 = 𝑢𝑣 − ∫𝑣 𝑑𝑢
Theorem 7
Trigonometri Integrals Rule
 ∫ 𝑠𝑖𝑛 𝑑𝑥 = −cos 𝑥 + 𝑐
 ∫ 𝑐𝑜𝑠 𝑑𝑥 = sin 𝑥 + 𝑐
 ∫
1
𝑐𝑜𝑠2 𝑥
𝑑𝑥 = tan 𝑥 + 𝑐
Where c is a constant
Prove Theorem 1
For prove theorem 1,we can differential 𝑥 𝑛+1
+ 𝑐 which be found at right space the
following ;
𝑑
𝑑𝑥
(𝑥 𝑛+1
+ 𝑐) = (𝑛 + 1)𝑥 𝑛
… . 𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑦 𝑡𝑤𝑜 𝑠𝑝𝑎𝑐𝑒 𝑤𝑖𝑡ℎ
1
𝑛 + 1
1
𝑛 + 1
.
𝑑
𝑑𝑥
(𝑥 𝑛+1
+ 𝑐) = (𝑛 + 1)𝑥 𝑛
.
1
𝑛 + 1
𝑑
𝑑𝑥
[
𝑥 𝑛+1
𝑛 + 1
+ 𝑐] = 𝑥 𝑛
So, ∫ 𝑥 𝑛
𝑑𝑥 =
1
𝑛+1
𝑥 𝑛+1
+ 𝑐
Prove Theorem 3 and 4
For prove theorem 4,we can differential ∫ 𝑓(𝑥)𝑑𝑥 ± ∫ 𝑔(𝑥)𝑑𝑥which be found at right
space the following ;
𝑑
𝑑𝑥
∫ 𝑓(𝑥)𝑑𝑥 ± ∫ 𝑔(𝑥)𝑑𝑥 =
𝑑
𝑑𝑥
[∫ 𝑓(𝑥)𝑑𝑥] ± [∫ 𝑔(𝑥)𝑑𝑥] = 𝑓(𝑥) ± 𝑔(𝑥)
𝑑
𝑑𝑥
∫ 𝑓(𝑥)𝑑𝑥 ± ∫ 𝑔(𝑥)𝑑𝑥 = 𝑓(𝑥) ± 𝑔(𝑥)
So,
∫ (𝑓(𝑥) ± 𝑔(𝑥))𝑑𝑥 = ∫ 𝑓(𝑥)𝑑𝑥 ± ∫ 𝑔(𝑥)𝑑𝑥
1. Find integral from ∫ (3𝑥2
− 3𝑥 + 7)𝑑𝑥!
Answers:
∫ (3𝑥2
− 3𝑥 + 7)𝑑𝑥 = 3∫ 𝑥2
𝑑𝑥 − 3∫ 𝑥 𝑑𝑥 + ∫ 7 𝑑𝑥 theorema 2,3 and 4
=
3
2+1
𝑥2+1
−
3
1+1
𝑥1+1
+ 7𝑥 + 𝑐 theorem 1

5. = 𝑥3
−
3
2
𝑥2
+ 7𝑥 + 𝑐
So, ∫ (3𝑥2
− 3𝑥 + 7)𝑑𝑥 = 𝑥3
−
3
2
𝑥2
+ 7𝑥 + 𝑐
Prove Theorem 6
In the class XI, you have know derivative of two times product of functions 𝑓(𝑥) =
𝑢(𝑥). 𝑣(𝑥) is
𝑑
𝑑𝑥
[𝑢(𝑥). 𝑣(𝑥)] = 𝑢(𝑥). 𝑣′(𝑥) + 𝑣(𝑥). 𝑣′
(𝑥)
It will prove that partial integral rule with formula them. Method them is with differential
two equation it the following :
∫
𝑑
𝑑𝑥
[𝑢(𝑥). 𝑣(𝑥)] = ∫ 𝑢(𝑥). 𝑣′(𝑥) + ∫ 𝑣(𝑥). 𝑣′
(𝑥)𝑑𝑥
𝑢(𝑥). 𝑣(𝑥) = ∫ 𝑢(𝑥). 𝑣′(𝑥) + ∫ 𝑣(𝑥). 𝑣′
(𝑥)dx
∫ 𝑢(𝑥). 𝑣′(𝑥) = 𝑢(𝑥). 𝑣(𝑥) − ∫ 𝑣(𝑥). 𝑣′
(𝑥)dx
Because v’(x) dx= dv and u’(x)dx=du
So,the equation can be written ∫ 𝑢 𝑑𝑣 = 𝑢𝑣 − ∫𝑣 𝑑𝑢
B.1 Substitution Integral Rule
Substitution Integral Rule is like which be written at Theorem 5. This rule was used
for to solve the problem in integration which not can to solve with base formulas what
already learn. For remainder it, example the following it
Example ;
1. Find the integral from
Answers;
a. Supposing that: u=9-x2
then du =-2x dx

6. So,
b. Supposing that u= √ 𝑥 =𝑥
1
2
with the result that
c. Supposing that u= 1- 2x2
and du = -4x dx
dx =
𝑑𝑢
−4𝑥
so the integral can be written the following
Substitution u= 1- 2x2
to equation 12u-3
+ c

7. So,
Prove theorem 7
In the class XI, you have learn derivative trigonometric function, is
𝑑
𝑑𝑥
(sin 𝑥) = cos 𝑥
𝑑
𝑑𝑥
(cos 𝑥) = −sin 𝑥 ,and
𝑑
𝑑𝑥
(tan 𝑥) = 𝑠𝑒𝑐2
𝑥
The following this we can prove trigonometric integral rule to use formulas. This method is
with integration two space the following;
From
𝑑
𝑑𝑥
(sin 𝑥) = cos 𝑥 be found∫ 𝑐𝑜𝑠 𝑑𝑥 = sin 𝑥 + 𝑐
From
𝑑
𝑑𝑥
(cos 𝑥) = −sin 𝑥 𝑏𝑒 𝑓𝑜𝑢𝑛𝑑 ∫ 𝑠𝑖𝑛 𝑑𝑥 = −cos 𝑥 + 𝑐
From
𝑑
𝑑𝑥
(tan 𝑥) = 𝑠𝑒𝑐2
𝑥 be found ∫ 𝑠𝑒𝑐2
𝑑𝑥 = tan 𝑥 + 𝑐
B.2 space integral with √𝒂 𝟐 − 𝒙 𝟐, √𝒂 𝟐 + 𝒙 𝟐 and √𝒙 𝟐 + 𝒂 𝟐
Integration spaces √𝑎2 − 𝑥2, √𝑎2 + 𝑥2 and √𝑥2 + 𝑎2 can be work with substitution with x =
a sin t, x= a tan t, x = a sec t. So can be found spaces the following it ;

8. Right angle for integral trigonometric substitution;
(𝑖)√𝑎2 − 𝑥2 = 𝑎 cos 𝑥, (𝑖𝑖)√𝑎2 + 𝑥2 = 𝑎 sec 𝑡, (𝑖𝑖𝑖)√𝑥2 − 𝑎2 =a tan x
1. Find each integral the following it:
Answers:
For to work this integral, you must change sin(3x+1)cos(3x+1) in the double angle
trigonometric formulas