Integration By Parts Formula

By Parts Formula Meaning

In calculus mathematics, Integration is considered a very important computation. We use a lot of rules and formulas to get the integration of some functions. There is a special rule that we know by the name as integration by parts. It is used for integrating the products of two functions. If you want to master the technique of integrations, I suggest, you use the integration by parts formula. A lot of times, a function is a product of other functions and therefore needs to be integrated. It is here that we may use the method of integration by parts i.e., to integrate the product of two functions. What we actually do here in this method is that we find the integrals by reducing them into standard forms.


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For example, lets us say that we have to find the integration of x sin x, so we will require the formula. We already know that integrand is actually the product of the two functions. So the integration by parts formula can be written as:


\[\int uv dx = udx - \int (\frac{du}{dx} \int v dx)dx\]


There are two more methods that we can use to perform the integration apart from the integration by parts formula,. They are:

  • The method of Integration by Substitution.

  • The method of Integration using Partial Fractions.

Ilate Rule Formula

The integration by parts formula taught us that we use the by parts formula when we are given the product of two functions. The ilate rule of integration considers the left term as the first function and the second term as the second function. We call this method ilate rule of integration or ilate rule formula.


For example, we are to integrate x ex so according to the ilate rule of integration, x will have to be considered as the first function and ex will have to be considered as the second function. So basically, while choosing the first function, we need to keep in mind that the choice of the function must make the integration of the derivative of the function easy.


Normally, for ilate rule formula,  functions such as Inverse, Algebraic, Logarithm, Trigonometric, Exponent are given more preferences.


Use of Integration by Parts Calculator

For the integration by parts formula, we can use a calculator. The steps to use the calculator is as follows:

Step 1: Start by entering the function in the input field.

Step 2: Next, click on the  “Evaluate the Integral” button to get the output.

Step 3: The integrated value will be displayed in the output field.


Integral Formulas Classification

All the formulas of integration can be classified into the following functions:

  1. Rational functions

  2. Irrational functions

  3. Trigonometric functions

  4. Inverse trigonometric functions

  5. Hyperbolic functions

  6. Inverse hyperbolic functions

  7. Exponential functions

  8. Logarithmic functions

  9. Gaussian functions

Fun Facts

  1. Ilate can be broken down alphabetically into or in a simple words we can say that Ilate stands for:

I: Inverse trigonometric functions i.e., arctan x, arcsec x, arcsin x etc.

L: Logarithmic functions i.e., ln x  , log5(x), etc.

A: Algebraic functions.

T: Trigonometric functions i.e., sinx, cosx, tan x, etc.

E: Exponential functions.

  1. For functions such as ∫ √x Sinx dx, we cannot use the integration by parts formula.

  2. While finding the second function integral, we do not add any constant.

Solved Example

Example 1) Find the integration by parts of \[\int log x^{2} dx\]


Solution 1) \[\int log x^{2} dx = 2 \int (log x) \cdot 1 dx\]


= \[2[log x \cdot \int 1 dx - \int (\frac{d}{dx}(log x) \int 1 dx) dx]\]


= \[2[(log x) \cdot x - \int \frac{1}{x} \cdot x dx] = 2[x log x - x] + c = 2x(log x - 1) + c\]


Example 2) Find the integration by parts of \[\int (log x)^{2} dx\]


Solution 2) \[\int (log x)^{2} dx = (log x)^{2} \int dx - \int [ \frac{d}{dx} (log x)^{2} \int dx] dx\]


= \[(log x)^{2} \cdot x - \int 2(log x) \frac{1}{x} \cdot x dx\]


= \[x(log x)^{2} - 2 \int log x dx\]


= \[x (log x)^{2} - 2x (log x - 1) + c\]


Example 3) Evaluate \[\int tan^{-1}x dx\]; hence, find the value of \[\int cot^{-1}x dx\].


Solution 3) \[\int tan^{-1}x dx = tan^{-1}x \int [ \frac{d}{dx} (tan^{-1}x) \int dx]dx\]


= \[(tan^{-1}x) \cdot x - \int \frac{1}{1 + x^{2}} \cdot x dx = x tan^{-1}x - \frac{1}{2} \int \frac{2x}{1 + x^{2}} dx\]


= \[x tan^{-1}x - \frac{1}{2} log (1 + x^{2}) + c\]


= \[\int cot^{-1} x dx = \int (\frac{\pi}{2} - tan^{-1}x) dx = \frac{\pi}{2} \int dx - \int tan^{-1} x dx\]


= \[\frac{\pi}{2} x - [x tan^{-1} x - \frac{1}{2} log (1 + x^{2})] + k\]


= \[x(\frac{\pi}{2} - tan^{-1}x) + \frac{1}{2} log(1 + x^{2}) + k\]


= \[c cot^{-1} x + \frac{1}{2} log(1 + x^{2}) + k\]

FAQ (Frequently Asked Questions)

Question 1) What is Integration?

Answer 1) The calculation of an integral is known as Integration. In maths, Integrals play a big role when we want to find useful quantities such as areas, volumes, displacement, etc. Speaking of integrals, we usually relate to definite integrals as the indefinite integrals are used for antiderivatives. In Mathematics, Integration is one of the two major calculus. The other being differentiation that is used to measure the rate of change of any function with respect to its variables. Integration is a vast topic that is usually covered in higher classes of 11 and 12.  So the basic job of integral calculation is to find the function that will explain the area, displacement, volume occurring due to a collection of small data that are incapable of being measured singularly.

Question 2) What are the Methods of Integration?

Answer 2) The different methods that we can use for integration are:

  1. Integration by Substitution method.

  2. Integration by Parts method.

  3. Integration Using Trigonometric Identities.

  4. Integration of Some particular function.

  5. Integration by partial fraction.

In the substitution method, we transform a given integral ∫ f(x) dx into another form. This can be done if we change the independent variable x to t i.e., by substituting x = g (t).


For the integration by parts formula, we find the integrals by reducing them into standard forms. Some integrals that include trigonometric functions can be solved using trigonometric identities.