Integral calculus concerned with the theory and applications of integrals and integration. Integral may be definite or indefinite but the process of solving either a definite or an indefinite integral is called integration. Integration is the reversal of the process of differentiating. The basic rules of integration are presented here along with several examples. The rules only apply when the integrals exist.


Sum Rule for Integrals

Let $f(x)$ and $g(x)$ be any two functions. The sum rule says that the integral of the sum of two functions is the sum of their separate integrals.

$\int [f(x)+g(x)] dx$ = $\int f(x) dx + \int g(x) dx$

$\int [f(x)-g(x)] dx$ = $\int f(x) dx -\int g(x) dx$

Example: $\int [x^{4}+ 2x^{2}]dx$ = $\int x^{4}+\int 2x^{2}dx$

= $\frac{x^{5}}{5}+\frac{2}{3}x^{3}$

Multiplication Rule for Integrals

$\int ax^{n}dx$ = $a\int x^{n}dx$ for any constant 'a'.

That is any constant factor may be taken outside the integration sign.

Example: $\int 4x^{3}dx$ = $4\int x^{3}dx$

= $4$$\frac{x^{4}}{4}$+c = $x^{4}+c$


$\int x^{n}dx$ = $\frac{x^{n+1}}{n+1}$+$c$ , Where $n\neq 0$

Here to integrate a power of $x$, increase the power of $x$ by one and divide the new power. If the power of $x$ be 'n' then the new power become (n+1).

For example, let $f(x)$ = $x^{4}$ be a function. The integral of the given function is given by $\int f(x) dx$ = $\int x^{4}dx$ = $\frac{x^{4+1}}{4+1}$ = $\frac{x^{5}}{5}$

This rule will not be used for the function $x^{-1}$. Because the integration of given function by the use of given rule will give undefined function.
All off the following expressions $(x^{3}+5)^{2}$, $e^{x^{2}}$, $\sin (x^{3})$ are the example of composite functions. To evaluate them for a particular value of x, you would need to work out the come off one function, and then use a second function to arrive at the final answer.

Example, for the expression $(x^{3}+5)^{2}$

$y$ = $(x^{3}+5)^{2}$ then the function can be broken down into two steps:

$u$ = $x^{3}+5$ and $y$ = $u^{2}$

The alternative way of writing the chain rule helps to understand the concept well,

If $y$ = $f(g(x))$ then $y$ = $f(u)$ and $u$ = $g(x)$

so $\frac{dy}{dx}$ = $\frac{dy}{du}$ $\times$ $\frac{du}{dx}$

= $f '(u) \times g'(x)$

= $g'(x).f'(g(x))$

Example: Evaluate the integral with respect to $x$, $\int 4(x^{2}+3x)^{3}(2x+3)dx$
Solution:
Here the derivative of ($x^{2}$+3x) is (2x+3).

Let $y$ = $(x^{2}+3x)^{4}$

We can split the above expression into

$y$ = $u^{4}$ and $u$ = $x^{2}+3x$

and $\frac{dy}{dx}$ = $4u^{3}$ = $4(x^{2}+3x)^{3}$

$\frac{du}{dx}$ = $2x+3$

By the chain rule,

$\frac{dy}{dx}$ = $\frac{dy}{du}$ $\times$ $\frac{du}{dx}$

$\frac{d}{dx}$ $(x^{2}+3x)^{4}$= $4(x^{2}+3x)^{3}(2x+3)$

By reversing the above expression, we have

$\int 4(x^{2}+3x)^{3}(2x+3)dx$ = $(x^{2}+3x)^{4}+c$

Integration by parts:

$\int u\ dv$ = $uv - \int v\ du$