The reverse operation of finding a derivative is called the antiderivative. A function $F$ is an antiderivative of a function  $f$  if  $F ’(x)$ =  $f (x)$. That antiderivative is nothing but the differentiation. Integration is the process of evaluating an indefinite integral or a definite integral.                   
$\int_a^b$ $f(x)dx$

The definite integral is a number whose value depends on the function $f$ and the numbers $a$ and $b$, and it is defined as the limit of a Riemann sum.
$f(x)$ = $\int$ $f'(x)dx$

$D_x\ [f(x)]$ = $f'(x)$

The indefinite integral  $\int$ $f’(x)dx$ is defined as a function $g$ such that its derivative $Dx[f(x)]$ = $f’(x)$. Indefinite integral involves an arbitrary constant, for example let us find the integral of $x^2$.

$\int$ $x^2 dx$ = $x^3 + c$

The arbitrary constant $c$ is called a constant of integration.

Constant of Integration is a constant that is added at the end of the result obtained by evaluating the indefinite integral of a given function. The derivative of a constant is $0$. But, when you integrate, you should consider that there is a constant involved, but we don’t know what it is for a particular function. Therefore, you can just use $C$ to represent the value. That arbitrary constant ‘$c$’ is called Constant of Integration. 
Every antiderivative $F$ of $f$ must be of the form $F(x)$ = $G(x) + C$, where $C$ is a constant. After finding the integration there will be a value ‘$a$’ at which the function value $f(a)$ will be given , by pludding that into our obtained integration result we can find the value of the constant ‘$C$’.
Constant of integration rule is that we need to add a constant at the end of the solution obtained after integration.

$\int$ $x^n dx$ = $\frac{x^{n + 1}}{n + 1}$ $+\ c$

$\int$ $kdx$ = $kc + c$ 
Example 1:

Find the constant $f$ integration if $f’(x)$ = $14x^6 + 12x^4 - 3x^2 + 7$ and $f(1)$ = $13$ ?

Solution:

We know that,

$f(x)$ = $\int$ $f' (x)dx$

Plug in $f’(x)$ in the above formula, we get

$f(x)$ = $\int$ $(14x^6 + 10x^4 - 3x^2 + 7)dx$

Now lets distribute the integral to all the terms of $f’(x)$, we get

$\int$ $(14x^6 + 10x^4 - 3x^2 + 7)dx$ = $\int$ $14x^6 dx +$ $\int$ $10x^4 dx -$ $\int$ $3x^2 dx +$ $\int$ $7dx$

Now lets do the integration of each term separately,

Here we will apply the power rule,

$\int$ $x^n\ dx$ = $\frac{x^{n+1}}{n + 1}$ $+ c$

$\int$ $14x^6 dx$ = $14$ $\int$ $x^6 dx$

= $14$ $\frac{x^{6 + 1}}{6 + 1}$

= $14$ $\frac{x^7}{7}$

= $\frac{14}{7}$ $x^7$

= $2x^7$

$\int$ $10x^4 dx$ = $10$ $\int$ $x^4 dx$

= $10$ $\frac{x^{4 + 1}}{4 + 1}$

= $10$ $\frac{x^5}{5}$

= $\frac{10}{5}$ $x^5$

= $2x^5$

  $\int$ $3x^2 dx$ = $3$ $\int$ $x^2 dx$

= $3$ $\frac{x^{2 + 1}}{2 + 1}$

= $3$ $\frac{x^3}{3}$

= $\frac{3}{3}$ $x^3$

= $x^3$

$\int$ $7dx$ = $7$ $\int$ $dx$

= $7x$

Now let us plug in the values obtained after integration,

$\int$ $(14x^6 + 10x^4 - 3x^2 + 7)dx$ = $\int$ $14x^6 dx +$ $\int$ $10x^4 dx -$ $\int$ $3x^2 dx +$ $\int$ $7dx$

= $2x^7 + 2x^5 - x^3 + 7x + C$

Here $C$ is the constant of integration.

$f(x)$ = $2x^7 + 2x^5 - x^3 + 7x + C$

Now lets plug in $x$ = $1$ in the above function, we get

$f(1)$ = $2(1^7) + 2(1^5) - (1^3) + 7(1) + C$

$f(1)$ = $2 + 2 - 1 + 7 + C$

Plug in $f(1)$ = $13$ in the above equation we get,

$13$ = $10 + C$

Subtracting $10$ on both sides, we get

$13 - 10$ = $10 - 10 + C$

$C$ = $3$

Now our constant of integration is $C$ = $3$.

Now our $f(x)$ will be,

$f(x)$ = $2x^7 + 2x^5 - x^3 + 7x + 3$

$\Rightarrow$ $\int$ $\frac{1}{x}$ $dx$ = $ln|x| + c$
Example 2: 

Find the constant of integration if $f’(x)$ = $3x - 1$ and $f(1)$ = $-3$ ?
  
Solution:

We know that,

$f(x)$ = $\int$ $f'(x)dx$

Plug in $f’(x)$ in the above formula, we get

$f(x)$ = $\int$ $3x^{-1} dx$

= $\int$ $3$ $\frac{1}{x}$ $dx$

= $3$ $\int$ $\frac{1}{x}$ $dx$

= $3ln|x| + c$

Here $C$ is the constant of integration.

$f(x)$ = $3ln|x| + c$

We know that,

$ln(1)$ = $0$

Plug in the value of $f(1)$ and $ln|1|$ in the equation, we get

$-3$ = $3(0) + c$

$-3$ = $0 + c$

$c$ = $-3$

Now our $f(x)$ will be,

$f(x)$ = $3ln|x|-3$

$\Rightarrow$ $\int$ $e^x dx$ = $e^x + C$
Example 3:

Find the constant of integration if $f’(x)$ = $-4e^x$ and $f(0)$ = $-32$ ?
  
Solution:

We know that,

$f(x)$ = $\int$ $f'(x)dx$

Plug in $f’(x)$ in the above formula, we get

$f(x)$ = $\int$ $-4e^x dx$

= $-4$ $\int$ $e^x dx$

$f(x)$ = $- 4e^x + c$

Here $C$ is the constant of integration.

Now plug in $x$ = $0$, we get

$f(0)$ = $-4e^0 + c$

We know that,

$e^0$ = $1$

Now plug in the value of $f(0)$ and $e^0$ in the above equation, we get

$-32$ = $-4(1) + c$

$-32$ = $-4 + c$

Adding $4$ on both sides,

$-32 + 4$ = $-4 + 4 + c$

$c$ = $-28$

Now our $f(x)$ will be,

$\Rightarrow$ $f(x)$ = $-4e^x - 28$