## Integration by parts formula

**Introduction:** Integration is an important part of mathematics and integration by parts was discovered by Brooke Taylor in 1715 which helped a lot in solving integration problems. ∫ is the integration sign and written at the beginning of any integration problem. Integration is used to calculate those functions for which differentiation exists.

But Sometimes an integration problem is very complex and can’t be solved as a whole. Therefore an additional technique of integration by parts helps in finding the integration of the products by partially integrating them.

## Common applications of Integration by parts formula

Integration by parts formula is used to calculate those integrals for which are common Integral formula does not exist. Common applications include inverse trigonometric functions, log functions, graphical representations, etc. having no holistic formula for integration.

## The formula for calculating Integration by parts:

To calculate an integral problem by integration by parts, we divide a function in the form of fx. gx. Thus we create a product rule of integration and then individually select them and then the solution is found individually.

Thus the formula is: ∫f(x).g(x).dx = f(x)∫ g(x).dx – ∫f’(x) ∫g(x).dx

Thus for integration in parts, you need to find out the Product of the first function and integration of the second function ( f(x)∫ g(x).dx). Then you have to subtract it from integration of differentiation of the first function, integration of the second function ∫f’(x) ∫g(x).dx. The result would be your answer.

## Step-by-step formula to calculate integration by parts:

First, we need to differentiate the function f(x)

Secondly, we will find the integration of the function g(x).

We’ll place these values in the formula and then derive the solution.

**Solved examples**

** Example no.1:** find the integration of x Cos(x) dx.

**Solution:** Here, the integral problem is a product of two functions: x and Cos x. So using the integration by parts formula, the solution is

∫f(x).g(x).dx = f(x)∫ g(x).dx – ∫f’(x) ∫g(x).dx

= x ∫Cos x dx – ∫differentiation of x ∫Cos x dx

= x Sin x – ∫1(sin x) dx

= x sin x – cosx +C (c represents constant).

**Ques 2:** What is the integration of ∫x. Ex dx?

**Ans:** Here, f(x)= x and g(x)= ex

So, derivative of the f(x)= 1 and integration of g(x) = ex

Putting in the formula ∫f(x).g(x).dx = f(x)∫ g(x).dx – ∫f’(x) ∫g(x).dx

=x (ex) – ∫1 (ex)

**Ans:** (x-1)ex +c