Integration along with differentiation are very important concepts within calculus. It requires a lot for someone to understand these concepts and get better. Integration has many types and there are different methods for doing integration. There are online integration calculators like this integral by parts calculator which allow you to perform different types of integration but one must learn the manual way of computing integration.

Since integration isn't a small concept, there are many sub topics of it. Washer method calculator along with disc method calculator and shell method calculator are some useful online calculator within this website which help you calculate these complex concepts and get step by step results along with plot and useful other visuals.

## What is Partial Fraction?

In integration, there are some functions which do not integrate into simple functions. For turning such functions into simpler functions, we use partial fraction. In partial fractions the integration will use logarithm only if the denominator is linear.

For doing integration by partial fraction, let’s evaluate an indefinite integral.

**Related:** For specifically dealing with indefinite integrals, use indefinite integrals calculator and this blog can help you a lot in understanding definite and indefinite integration.

$ \int \frac{x-5}{(2x-3)(x-1)}dx {2}lt;/p>

Now in attempting this, we might try to say, all right, is the numerator here the derivative or a constant multiple of the derivative of the denominator?

In which case, u-substitution might apply, but it's not the case here.

So what do we do?

## Using Partial Fraction Decomposition

Partial fraction decomposition might invoke some memories from a precalculus class or maybe from an algebra 2 class, but it's a technique to break up this rational expression into the sum of two rational expressions. In the equation above, we have a denominator which is factorable into (2x-3)(x-1) two expressions.

Before starting with partial fraction decomposition, there is a question!

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## Can we express a given fraction under the integral as a sum of two rational expressions?

The answer to that question is yes we can express a given fraction under the integral as a sum of two rational expressions. It is written as:

$ \frac{x-5}{(2x-3)(x-1)} = \frac{A}{2x-3} + \frac{B}{x-1} {2}lt;/p>

Now remember that it happens by simply using partial fraction decomposition rules. But keep in mind that the partial fraction is applicable to those fractions, where degree of polynomial in denominator is greater than or equal to degree of polynomial in numerator.

**Related:** Evaluate definite integral calculator and get step by step results.

## Partial Fraction Rules

Here are all basic rules for decomposition of fraction into partial fraction.

Factors in Denominator | Term in Partial Fraction Decomposition |

$ ax+b{2}lt;/td> | $ \frac{A}{ax+b}{2}lt;/td> |

$ (ax+b)^2{2}lt;/td> | $ \frac{A_1}{ax+b} + \frac{A_2}{(ax+b)^2} + ... + \frac{A_k}{(ax+b)^k} \;\;\; ,k=1,2,3,...{2}lt;/td> |

$ ax^2+bx+c {2}lt;/td> | $ \frac{Ax+B}{ax^2+bx+c}{2}lt;/td> |

$ (ax^2+bx+c)^2 {2}lt;/td> | $ \frac{A_1x+B_1}{ax^2+bx+c} + \frac{A_2 x+B_2}{(ax^2+bx+c)^2} + ... + \frac{A_k x+B_k}{(ax^2+bx+c)^k} \;\;\; ,k=1,2,3,...{2}lt;/td> |

So, let’s move back to our fraction, we have to solve A and B but it is not calculus, it is basically a precalculus or algebra which we learned at school level.

## How to do Integration by Partial Fraction

Let’s start doing integration by partial fraction. We can write,

$ \frac{A}{2x-3} + \frac{B}{x-1} = \frac{(A+2B)x \; - \; (A+3B) }{(2x-3)(x-1)} {2}lt;/p>

Also,

$ \frac{x-5}{(2x-3)(x-1)} = \frac{(A+2B)x \; - \; (A+3B) }{(2x-3)(x-1)} {2}lt;/p>

**Related:** This website helps you to calculate laplase transformation and you can also find online calculator for calculating fourier transformation online.

## Compare Coefficients in Numerator

We have values both on the right hand side and the left hand side. Now, we will compare coefficients in numerator from both L.H.S and R.H.S

**By coefficient of "x"** ⇒ A+2B=1

**By coefficient of "constant"** ⇒ A+3B=5

If we subtract these both equations then it give us: $B=4{2}lt;/p>

And then substitute the value of B in any upper equation

So,

$A+2B=1$ $A+2(4)=1$ $⇒A=-7{2}lt;/p>

Now, we can write

$ \frac{x-5}{(2x-3)(x-1)} = \frac{-7}{2x-3} + \frac{4}{x-1} {2}lt;/p>

So just like that, we can rewrite this entire integral.

$ \int \frac{x-5}{(2x-3)(x-1)}dx = \int \left( \frac{-7}{2x-3} + \frac{4}{x-1} \right) dx{2}lt;/p>

Now the above equation has simplified the whole integral as we have seen integrals like this before. This is going to be the same thing as,

$ \int \left( \frac{-7}{2x-3} \frac{4}{x-1} \right) dx = \frac{-7}{2} \int \frac{2}{2x-3}dx \;+\; 4 \int \frac{4}{x-1}dx{2}lt;/p>

Now, it's become quite simple to do as there are functions in denominators of both integrals which contain their respective derivatives at numerator.

**Related:** How to find the Volume of Solid of Revolution?

## Apply Natural Log

So, we have to apply natural log rule here simply,

$ \int \left( \frac{-7}{2x-3} \frac{4}{x-1} \right) dx = \frac{-7}{2} ln|2x-3| \;+\; 4 ln|x-1| \;+\; c{2}lt;/p>

This is integrated. We used the old concepts from the algebra and made the integral easy to solve and succeeded. The following procedure can easily perform integration by partial fraction.