Hello students, I am Bijoy Sir and welcome to our educational forum or portal.
Today we will discuss about the Integration, but you of all know that very well, Integration is a huge part in mathematics. Then all of the topics of Integration can’t be covered in one class, so we will discuss the all topics but lightly, means we will see all types of Integration in with the some example. We will take more illustration on the topic of Integration in upcoming classes or tutorials.
Let’s start the discussion –

Now take a heading – Integration

– Students this is the class of Integration so we will start from very beginning. Before starting Integration I want to tell you something that, Integration mainly developed by Newton Sir.

History of Integration

When Newton sir was trying to calculate the value gravitational potential, but the same time had no sufficient formula for proper calculation. Then Newton sir was thinking we need a new type of formula, that can give us a new way to sum. Then Newton sir had given a way to adding some element, which known as summation or Integration. We can add the very small element by the Integration. Which we will discuss very well in the tutorials of Definite Integration or Summation Series

Definition of Integration –

In the earlier section we already learnt the derivatives or differentiation, foe properly understanding the Integration we have to just remind the derivatives or differentiation.
Definition- In generally Integration is the inverse process of derivatives or in simple language integration is anti derivative s or primitive of derivatives.
In mathematically,
Let the derivative of a function [ f(x)] is g(x). Then,

\( \frac {df(x)}{dx}= g(x) \)

Then we can say Integration of g(x) is equal to f(x). [ where derivative of f(x) is g(x).]

So, we can write ,

\( \frac{df(x)}{dx}=g(x) \)

Or, d f(x) = g(x) ·dx

                                                           or, f(x) = ​\( \frac{1}{d} \)​g(x)· dx

Hence,  f(x) = ∫ g(x)· dx + c  [ where ‘c’ is integral constant ]

Note: If we take limit on integration the constant term ‘c’ will be automatically vanished.
# We can write that equation in another way

like, ​\( \frac {df(x)}{dx}=g(x) \)

Or, d f(x) = g(x) . dx

Now taking the integration on the both side we get,

or, ∫ d f(x) = ∫ g(x)· dx + c

or, ​\( \frac{1}{d} \) · df(x) = ∫ g(x)· dx  + c

∴  f(x) = ∫ g(x)· dx   + c

[ where, ∫ = 1/d we already learnt that ]

Now it’s time to see the illustration or time to learning the integration, that how can we deal with the integration.

Now we discuss the types of integration

Integration is two types – (a) Indefinite Integration and (b) Definite Integration. We study both of them but in separate classes.

 Indefinite Integration :

Indefinite Integration mainly use learning the integration that we can apply in Definite integration or in higher studies.
Indefinite integration divides in three types according to the solving method – i) Basic integration ii) By substitution, iii) By parts method, and another part is integration on some special function.

Downlad Here Integration Formula In Pdf File

i) Basic Integration :
Basic integration is simplest type of integration comparison to other types of integration. We just do it by separation the function, means in this case first of all we convert the given function in a simple form and then do it.
Let try to understanding by some illustration.

Illustration :1

Evaluate the given integral ​\( \int \frac{1}{x^(\frac{-2}{3})} dx \)

This is the most of basic integration like,

  ∫ xn dx =\( \frac {x^(n+1)}{n+1} \)​+ c

Here n =   +2/3, hence the required integration

or, \( \int \frac{1}{x^(\frac{-2}{3})} dx \)​=​\( \frac{x^(\frac{2}{3}+1)}{(\frac{2}{3}+1)} \)​+c

or, \( \int \frac{1}{x^(\frac{-2}{3})} dx \)​ = ​\( \frac{3}{5}x^\frac{5}{3} \)​ + c

Illustration :2

Evaluate the integration of ∫ cos2x · dx or integration of cos^2x

Ans:                                                       Let,  I = ∫ cos2 x · dx

or,  I = 1/2 [∫ 2·cos2x dx]

or,   I = [ ∫ (1 + cos2x) dx]

or, 2I = ∫ dx + ∫ cos2x dx +c

or,  2I =  x + sin2x/2 + c

or, I = ​\( \frac{1}{2}[ x + \frac{sin2x}{2}] \)​ + c

II) Substitution Method :
Definition of substitution method – Integration is made easier with the help of substitution on various variables. We will see a function will be simple by substitution for the given variable. This is one of the most important and useful methods for evaluating the integral.
Now we will take few examples that help us to understanding the concepts.

Example :1

Evaluate the integration of ​\( \int \frac{1}{x (lnx)^n} dx \)​, n ≠ 1

Ans:                                                      Let,  I = \( \int \frac{1}{x (lnx)^n} dx \)

[Now we are trying to understand that how to use substitution methods in integral]
Now, we substitute the (lnx) by replacing ‘z’
Hence, lnx = z
Now differentiate on both sides with respect to ‘x’ we get,

or,  ​\( \frac{d(lnx)}{dx} = \frac{dz}{dx} \)

or, (1/x)· dx = dz

And now putting the value of 1/x and( lnx) we get,


Hence the required integration is equal to ​\( \frac{(lnx)^(-n+1)}{-n+1} \)​ + c

Example :2

Evaluate the integration of


Now we will replace the xn – 1 by z2, hence xn – 1 = z2

And now differentiate both sides with respect to ‘x’ we get,

or, ​\( \frac {d(x^n-1)}{dx} = \frac{dz^2}{dx} \)

or, n · xn – 1 dx = 2 z dz

or, n · ​\( \frac {x^n}{x} \)​· dx = 2 z dz

Now putting the value of xn , from xn – 1 = z2, xn = z2 + 1

We get,                                                      or, n·​\( \frac {z^2 + 1}{x}dx = 2zdz \)

or, ​\( \frac{dx}{x} = \frac{2z} {n(z^2+1)}dz \)

And now putting the value of dx/x and xn-1 in the given integral we get,



III) By parts methods of Integration :
When the integration has given in the form of a product of two functions, in this case the integral can’t be able to solve by transformation or substitution methods, then we use this method to solve the given integral.
# Fundamental rules of By parts Methods –
In this case we use, a product of two functions f(x) and g(x) . So now our Integral looks like in the form = ∫ f(x) · g(x) dx .

* The fundamental formula of By Parts Methods –

I = ∫ f(x) · g(x) dx  = f(x) ∫ g(x) dx – ​\( \int[\frac{df(x)}{dx}\int g(x)dx] dx \)

This is the main fundamental rule to solving by parts methods.
But, there has some problems or confusion that which function we take first f(x) or g(x), to clear this problem, we have a solution that also fundamental rule is the form of ILATE or LIATE, we take the function first according to this name, as which come first.
So, let’s see some example to better understanding.

Example :1

Evaluate the integral ∫ x2 ex dx

Ans:                                   Let,  I = ∫ x2 ex dx and x2 = f(x)  & ex = g(x)

Then, according to the rule we have to take f(x) first due to algebraic function.

Given I = ∫ x2 ex   now write the fundamental formula we get,


In the second part, we have to use by parts once again due to product of two functions.



Note: Students, a full class on by parts methods will be coming soon, so there we will discuss briefly about the by parts methods. Then please subscribe our website [

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