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Partial Fractions

Learn definition of partial fractions with examples and cases. Practice partial fraction decomposition of the function & integration substitution.

Partial Fraction Decomposition

Partial Fraction

  • A rational function which is a ratio of two polynomial functions can be expressed as a sum of two or more simplex functions.
  • These simplex functions are called its partial fractions.
  • These partial fractions are easy to integrate.

Partial Fraction decomposition

  • The process of breaking a rational function i.e., \(\left(\dfrac{P(x)}{Q(x)}\right)\) into partial fractions is known as partial fraction decomposition.

For example:

\(\dfrac {1}{x+3} - \dfrac{2}{x+1}=\dfrac{(x+1)-2(x+3)}{(x+3)(x+1)}\)

\(\implies \)\(\dfrac {1}{x+3} - \dfrac{2}{x+1}=\dfrac{(-x-5)}{x^2+4x+3}\)

  • The method of obtaining L.H.S. from R.H.S. is called finding partial fractions.
  • It can be seen that the partial fractions on L.H.S. are easy to integrate.

\(\displaystyle\int \dfrac{-x-5}{x^2+4x+3}dx=\int\dfrac{1}{x+3}dx-\int\dfrac{2}{x+1}dx\)

\(=\ell n |x+3|-2\ell n|x+1|+C\)

  •  A rational function \(\left(\dfrac{P(x)}{Q(x)}\right)\)can have many forms of the polynomial \(Q(x)\).
  • Depending on the form of polynomial \(Q(x)\), we have several cases:
  1. When the denominator \(Q(x)\) is a product of distinct linear factors.
  2. When the denominator \(Q(x)\) contains irreducible linear quadratic factors.
  3. When the denominator \(Q(x)\) is a product of repeated linear factors.

Here, we are discussing the case, 

When denominator \(Q(x)\) is a product of distinct linear factors

  • Consider a rational function \(f(x)=\dfrac{P(x)}{Q(x)}\) where P and Q are polynomials and degree P < degree Q, such rational functions are called proper.
  • If \(Q(x)=(a_1x+b_1)(a_2x+b_2).....(a_nx+b_n)\) where no factor is repeated, then

\(\dfrac{P(x)}{Q(x)}=\dfrac{A_1}{a_1x+b_1}+\dfrac{A_2}{a_2x+b_2}+......+\dfrac{A_n}{a_nx+b_n}\)

where \(A_1,A_2,A_3....A_n\)  are constants and can be determined.

  • These are called partial fractions coefficients.

For example:

\(1.\;\dfrac{2}{x(x-1)}\) ; partial fractions are \(\dfrac{A}{x}+\dfrac{B}{x-1}\)

\(2. \dfrac{2x+3}{(x+1)(x+2)}\); partial fractions are \(\dfrac{A}{x+1}+\dfrac{B}{x+2}\)

\(3. \dfrac{3x+5}{(x-a)(x-b)(x-c)}\); partial fractions are \(\dfrac{A}{(x-a)}+\dfrac{B}{(x-b)}+\dfrac{C}{(x-c)}\)

Case 3

\(Q(x)\) contains linear as well as irreducible quadratic factors which are distinct.

If \(Q(x)\) has factor \(ax^2+bx+c\) (where \(b^2-4ac<0\)) then the expression \(\dfrac{P(x)}{Q(x)}\) will have a factor of form \(\dfrac{Ax+B}{ax^2+bx+c}\).

Where \(A,B\) are partial fraction coefficient.

\(2.\;\dfrac{2x-3}{(x+1)(x+2)}\); partial fractions are \(\dfrac{A}{x+1}+\dfrac{B}{x+2}\)

\(3.\;\dfrac{3x+5}{(x-a)(x-b)(x-c)}\); partial fractions are \(\dfrac{A}{x-a}+\dfrac{B}{x-b}+\dfrac{C}{(x-c)}\)

 

 

Illustration Questions

Write the form of partial fraction decomposition of the function. \(f(x)=\dfrac{2+3x}{(x-3)(2x+1)}\)

A \(f(x)=\dfrac{A_1}{(3x+2)}+\dfrac{A_2}{(x-2)}\)

B \(f(x)=\dfrac{A_1}{(5x-9)}-\dfrac{A_2}{(7x+3)}\)

C \(f(x)=\dfrac{A_1}{(x-3)}+\dfrac{A_2}{(2x+1)}\)

D \(f(x)=\dfrac{A_1}{(x+7)}+\dfrac{A_2}{(2x+3)}\)

×

When \(Q(x)\) has distinct linear factors

\(\dfrac{P(x)}{Q(x)}=\dfrac{A_1}{a_1x+b_1}+\dfrac{A_2}{a_2x+b_2}+......+\dfrac{A_n}{a_nx+b_n}\)

\(\therefore\) In this case,

\(\dfrac{2+3x}{(x-3)(2x+1)}=\dfrac{A_1}{x-3} + \dfrac{A_2}{2x+1}\)

R.H.S. is the partial fraction decomposition.

Write the form of partial fraction decomposition of the function. \(f(x)=\dfrac{2+3x}{(x-3)(2x+1)}\)

A

\(f(x)=\dfrac{A_1}{(3x+2)}+\dfrac{A_2}{(x-2)}\)

.

B

\(f(x)=\dfrac{A_1}{(5x-9)}-\dfrac{A_2}{(7x+3)}\)

C

\(f(x)=\dfrac{A_1}{(x-3)}+\dfrac{A_2}{(2x+1)}\)

D

\(f(x)=\dfrac{A_1}{(x+7)}+\dfrac{A_2}{(2x+3)}\)

Option C is Correct

Illustration Questions

Write the form of partial fraction decomposition of the function. \(f(x)=\dfrac{3x+5}{(x-1)^3(x-2)}\)

A \(f(x)=\dfrac{A}{(x-1)^2}+\dfrac{B}{(x+5)}+\dfrac{C}{(x-6)}\)

B \(f(x)=\dfrac{A}{(x-1)}+\dfrac{B}{(x-1)^2}+\dfrac{C}{(x-1)^3}+ \dfrac{D}{(x-2)}\)

C \(f(x)=\dfrac{A}{(x-1)^2}+\dfrac{B}{x}+\dfrac{C}{x-1}\)

D \(f(x)=\dfrac{A}{x-12}+\dfrac{B}{3x+5}+\dfrac{C}{x}\)

×

When \(Q(x)\) has repeated linear factors, \(\dfrac{P(x)}{Q(x)}\) will have the partial fraction of the form in which factor is repeated r times, will have r terms.

\(\therefore\) In this case,

\(\dfrac{3x+5}{(x-1)^3(x-2)}=\dfrac{A}{(x-1)}+\dfrac{B}{(x-1)^2}+\dfrac{C}{(x-1)^3}+ \dfrac{D}{(x-2)}\)

R.H.S is the partial fraction decomposition (3 factors for \((x-1)^3\))

Write the form of partial fraction decomposition of the function. \(f(x)=\dfrac{3x+5}{(x-1)^3(x-2)}\)

A

\(f(x)=\dfrac{A}{(x-1)^2}+\dfrac{B}{(x+5)}+\dfrac{C}{(x-6)}\)

.

B

\(f(x)=\dfrac{A}{(x-1)}+\dfrac{B}{(x-1)^2}+\dfrac{C}{(x-1)^3}+ \dfrac{D}{(x-2)}\)

C

\(f(x)=\dfrac{A}{(x-1)^2}+\dfrac{B}{x}+\dfrac{C}{x-1}\)

D

\(f(x)=\dfrac{A}{x-12}+\dfrac{B}{3x+5}+\dfrac{C}{x}\)

Option B is Correct

Illustration Questions

Write the form of partial fraction decomposition of the function. \(f(x)=\dfrac{2x^2-5x+1}{(x^2+1)(x-7)}\).

A \(f(x)=\dfrac{A}{x^2+1}+\dfrac{B}{x-7}\)

B \(f(x)=\dfrac{A}{5x-3}+\dfrac{Bx+C}{x+1}\)

C \(f(x)=\dfrac{Ax+B}{x^2+1}+\dfrac{C}{x-7}\)

D \(f(x)=\dfrac{Ax}{5x+1}+\dfrac{C}{x+1}\)

×

\(\dfrac{P(x)}{Q(x)}=f(x)=\dfrac{2x^2-5x+1}{(x^2+1)(x-7)}\)

The \(x^2+a^2\) factor of \(Q(x)\) will have factor of the form \(\dfrac{Ax+B}{x^2+a^2}\) in partial fraction.

\(\therefore f(x)=\dfrac{Ax+B}{x^2+1}+\dfrac{C}{x-7}\)

R.H.S. is the partial fraction decomposition.

Write the form of partial fraction decomposition of the function. \(f(x)=\dfrac{2x^2-5x+1}{(x^2+1)(x-7)}\).

A

\(f(x)=\dfrac{A}{x^2+1}+\dfrac{B}{x-7}\)

.

B

\(f(x)=\dfrac{A}{5x-3}+\dfrac{Bx+C}{x+1}\)

C

\(f(x)=\dfrac{Ax+B}{x^2+1}+\dfrac{C}{x-7}\)

D

\(f(x)=\dfrac{Ax}{5x+1}+\dfrac{C}{x+1}\)

Option C is Correct

Partial Fraction Decomposition of P(x)/Q(x) Form (where Degree of P(x) < Degree of Q(x))

  • Consider a rational function of the form \(\dfrac{P(x)}{Q(x)}\) where P(x) and Q(x) are polynomial in \(x\) and degree of P(x) < degree of Q(x).
  • \(Q(x)\) can occur in combination of linear and quadratic function or repeated quadratic function, etc.
  • Here, we will discuss the case in which \(Q(x)\) occurs in form of repeated quadratic function.
  • If \(Q(x)\) contains repeated quadratic function that cannot be reduced to linear factors, then the partial fraction will have as many terms as it is repeated in partial fraction decomposition.

Case: When \(Q(x)\) contains repeated quadratic function

Let \(Q(x)\) is of the form  \(\dfrac{P(x)}{(ax^2+bx+c)^2}\) and \(b^2-4ac<0\)  

Here, combination of two cases is used, one is quadratic function and the other one is repetition of function.

 

For quadratic function we use \(\dfrac{1}{ax^2+bx+c}\Rightarrow\;\dfrac{A_1x+B_1}{ax^2+bx+c}\)

For repeatition of function, we use

\(\dfrac{1}{(x+a)^3}\Rightarrow\;\dfrac{A_1}{x+a}+\dfrac{A_2}{(x+a)^2}+\dfrac{A_3}{(x+a)^3}\)

Combining these both, we can write the partial factors decomposition for \(\dfrac{P(x)}{(ax^2+bx+c)^2}\)

as \(\dfrac{A_1x+B_1}{(ax^2+bx+c)}+\dfrac{A_2x+B_2}{(ax^2+bx+c)^2}\)

  • For \(n^{th}\) power of \(ax^2+bx+c\),

\(\dfrac{P(x)}{(ax^2+bx+c)^n}\Rightarrow\;\dfrac{A_1x+B_1}{(ax^2+bx+c)}+\dfrac{A_2x+B_2}{(ax^2+bx+c)^2}+\,.....\,\dfrac{A_nx+B_n}{(ax^2+bx+c)^n}\)

For example:

\(1.\;f(x)=\dfrac{2x}{(x-3)(x^2+2x+1)^2}\)

Then, its partial fractions are

\(f(x)=\dfrac{A_1}{(x-3)}+\dfrac{A_2x+B_1}{(x^2+2x+1)}+\dfrac{A_3x+B_2}{(x^2+2x+1)^2}\)

\(2.\;f(x)=\dfrac{x^2-2x+1}{(x^2+3x+1)^2\,(x+3)^2}\)

Then, its partial fractions are

\(f(x)=\dfrac{A_1x+B_1}{(x^2+3x+1)}+\dfrac{A_2x+B_2}{(x^2+3x+1)^2}+\dfrac{A_3}{x+3}+\dfrac{A_4}{(x+3)^2}\)

Illustration Questions

Write the form of partial fraction decomposition of \(f(x)=\dfrac{2x-3}{(x+1)(x^2+x+2)^2} \)

A \(f(x)=\dfrac{A}{x+1}+\dfrac{B}{x^2+x+2}+\dfrac{C}{(x^2+x+2)^2} \)

B \(f(x)=\dfrac{A}{x+1}+\dfrac{Bx+C}{x^2+x+2}+\dfrac{Dx+E}{(x^2+x+2)^2} \)

C \(f(x)=\dfrac{A}{(x+1)^2}+\dfrac{Bx+C}{(x^2+x+1)}\)

D \(f(x)=\dfrac{A}{(x^2+x+1)^2}-\dfrac{B}{x+1}\)

×

In \(\dfrac{P(x)}{Q(x)}\) when \(Q(x)\) has irreducible quadratic repeated factor then, that factor will have as many terms as it is repeated in partial fraction decomposition.

In this case,

\(\dfrac{P(x)}{Q(x)}=f(x)=\dfrac{2x-3}{(x+1)(x^2+x+2)^2}\)

\(=\dfrac{A}{x+1}+\dfrac{Bx+C}{x^2+x+2}+\dfrac{Dx+E}{(x^2+x+2)^2}\)

Write the form of partial fraction decomposition of \(f(x)=\dfrac{2x-3}{(x+1)(x^2+x+2)^2} \)

A

\(f(x)=\dfrac{A}{x+1}+\dfrac{B}{x^2+x+2}+\dfrac{C}{(x^2+x+2)^2} \)

.

B

\(f(x)=\dfrac{A}{x+1}+\dfrac{Bx+C}{x^2+x+2}+\dfrac{Dx+E}{(x^2+x+2)^2} \)

C

\(f(x)=\dfrac{A}{(x+1)^2}+\dfrac{Bx+C}{(x^2+x+1)}\)

D

\(f(x)=\dfrac{A}{(x^2+x+1)^2}-\dfrac{B}{x+1}\)

Option B is Correct

Decomposition of the Partial Fraction of the Form of \(\dfrac{P(x)}{Q(x)}\), where \(P(x)\) and \(Q(x)\) are Polynomial in \(x\)  and Degree \(P(x)\geq{\text{degree}}\,\, Q(x)\)

  • To decompose \(\dfrac{P(x)}{Q(x)}\)  ,

         (1) Firstly, we divide \(P(x)\) by \(Q(x)\) 

         e.g  Let  \(I= \dfrac{3x^3+11\,x^2+9x+6}{x^2+2x+3}\)

Where,  \(P(x) = 3x^3+11\,x^2+9\,x +6\) 

    and   \(Q(x)=x^2+2\,x+3\)

(2)  Now, separating the quotient \([P(x)]\) from actual fraction and dividing \(P(x)\) by \(Q(x)\)

\(\dfrac{P(x)}{Q(x)} = \dfrac{3x^3+11\,x^2+9x+6}{x^2+2x+3}\)

 

Here , Quotient  \(=3x+5\) 

Divisor \( = x^2+2x+3\) 

Dividend \(=3\,x^3+11\,x^2+9\,x+6\) 

Remainder  \(=-10\,x-9\) 

 

Now, we will use the relation among dividend, Quotient, Divisor and Remainder.

Dividend = Quotient × Divisor + Remainder 

\(3x^3+11x^2+9x+6=(3x+5)(x^2+2x+3)+(-10x-9)\)

\(\dfrac{P(x)}{Q(x)}=\text{Quotient}+\dfrac{\text{Remainder}}{\text{Divisor}}\)

So, \(I=3x+5+\dfrac{(10x-9)}{x^2+2x+3}\)

\(I=3x+5-\dfrac{10x+9}{x^2+2x+3}\)

Here, \(\dfrac{10x+9}{x^2+2x+3}\)

For example:

\(1.\;\dfrac{P(x)}{Q(x)}=\dfrac{x^3-4x-10}{x^2-x-6}=x+1+\dfrac{3x-4}{x^2-x-6}\)

\(2.\;\dfrac{P(x)}{Q(x)}=\dfrac{2x^3-11x^2-2x+2}{2x^2+x-1}=(x-6)+\dfrac{5x-4}{2x^2+x-1}\)

Illustration Questions

Write the form of partial fraction decomposition of the function . \(I= \dfrac{2\,x^3+3\,x^2+x+5}{x^2+2\,x+1}\)

A \((2x-5) +\dfrac{3}{x+1} +\dfrac{9}{(x+1)^2}\)

B \((2x-1) +\dfrac{1}{x+1} +\dfrac{5}{(x+1)^2}\)

C \((3\,x-2) +\dfrac{1}{y+1} +\dfrac{4}{x+2}\)

D \((9\,x-1) +\dfrac{1}{(2x+1)} +\dfrac{1}{(3x+1)}\)

×

Here,  \(P(x)=2\,x^3+3\,x^2 +x+5\) 

and  \(Q(x) = x^2+2x+1\)

Now , we divide \(P(x) \) by \(Q(x) \) 

\(\dfrac{P(x)}{Q(x)}= \dfrac{2\,x^3+3\,x^2+x+5}{x^2+2\,x+1}\)

 

image

Here,

Quotient \(= 2\,x-1\),

Reminder = \(x +6\) ,

Divisor  \(=x^2+2\,x+1\),

Dividend \(= 2\,x^3+3\,x^2 +x+5\)

We know that,

Dividend = Quotient × Divisor + Remainder 

\(= (x^2+2\,x+1) (2\,x-1) +(x+6)\)   ........(1)

Now, we replace \(P(x)\) by \((1)\)in the given fraction.

\(I = \dfrac{(x^2+2\,x+1)(2\,x-1)+(x+6)}{x^2+2\,x+1}\)

\( = (2\,x-1) + \dfrac{(x+6)}{x^2+2\,x+1}\)   ........(2)

Decompose  \(x^2+2\,x+1\)  into its factors 

\(\Rightarrow x^2+2\,x+1= (x+1)^2 = (x+1) (x+1)\)   ......(3)

Replace  \(x^2+2\,x+1\)  by \((x+1)^2\) 

\(\Rightarrow (2x-1) +\dfrac{x+6}{(x+1)^2}\)    ......(4)

Now decompose \(\dfrac{x+6}{(x+1)^2}\) by using  partial fraction .

\(\Rightarrow\dfrac{x+6}{(x+1)(x+1)} = \dfrac{A}{(x+1)} +\dfrac{B}{(x+1)^2}\)

\(\Rightarrow \dfrac{x+6}{(x+1)^2} = \dfrac{A(x+1)+B}{(x+1)^2}\)

\(= x+6 = A(x+1)+B\)

\(=x+6=A\,x+A+B\)

Now we compare coefficients of both side,

i.e., \(A\,x=x,\)  \(A+B=6\) 

we get,

\(A=1\),   \(B=5\)

Put the values of A and B in equation (4)

So, the decomposed form of the fraction  is :- 

\(\Rightarrow(2\,x-1) +\dfrac{1}{x+1} + \dfrac{5}{(x+1)^2}\)

Hence , Option (B) is correct.

Write the form of partial fraction decomposition of the function . \(I= \dfrac{2\,x^3+3\,x^2+x+5}{x^2+2\,x+1}\)

A

\((2x-5) +\dfrac{3}{x+1} +\dfrac{9}{(x+1)^2}\)

.

B

\((2x-1) +\dfrac{1}{x+1} +\dfrac{5}{(x+1)^2}\)

C

\((3\,x-2) +\dfrac{1}{y+1} +\dfrac{4}{x+2}\)

D

\((9\,x-1) +\dfrac{1}{(2x+1)} +\dfrac{1}{(3x+1)}\)

Option B is Correct

Trick for Simplifying the Function

  • Sometimes, we come across the functions which are a bit difficult to integrate.
  • Such functions can't be simplified using partial fraction decomposition.
  • To integrate them easily, we simplify them using algebraic operations.
  • Consider some examples to understand this.

\(1.\;\displaystyle\int\dfrac {t^2}{t^2-2}\;dt\)

It seems quite difficult.

  • To make it easy, we will add and subtract 2 in numerator.

\(\displaystyle\int\dfrac {t^2-2+2}{t^2-2}\;dt\)

Now, separate \(t^2-2\;\&\;+2\),

\(\displaystyle\int\left(\dfrac {t^2-2}{t^2-2}+\dfrac{2}{t^2-2}\right)\;dt\)

\(=\displaystyle\int \left ( 1+\dfrac {2}{t^2-2}\right)\;dt\)

\(2.\;\displaystyle\int\dfrac {x}{x-1}\;dx\)

To make this easy, we will add and subtract 1.

\(\displaystyle\int\dfrac {x-1+1}{x-1}\;dx\)

\(=\displaystyle\int\left(\dfrac {x-1}{x-1}+\dfrac{1}{x-1}\right)\;dx\)

\(=\displaystyle\int\left(1+\dfrac{1}{x-1}\right)\;dx\)

This can be integrated easily.

Illustration Questions

Which one of the following operations will be suitable for integration of  \(\displaystyle\int\dfrac {t}{t-2}\;dt\)

A \(\displaystyle\int\dfrac {t-2+2}{t-2}\;dt\)

B \(\displaystyle\int\dfrac {t×t}{(t-2)\,t}\;dt\)

C \(\displaystyle\int \left (\dfrac {t}{t-2}×\dfrac {t-2}{t-2}\;\right)dt\)

D \(\displaystyle\int \left (\dfrac {t}{t-2}×\dfrac {2}{2}\;\right)dt\)

×

\(I=\displaystyle\int\dfrac {t}{t-2}\;dt\)

\(\Rightarrow\displaystyle\int\dfrac {t-2+2}{t-2}\;dt\)

OR

\(\Rightarrow\displaystyle\int \left [1+\dfrac {2}{t-2}\right]\;dt\)

Which one of the following operations will be suitable for integration of  \(\displaystyle\int\dfrac {t}{t-2}\;dt\)

A

\(\displaystyle\int\dfrac {t-2+2}{t-2}\;dt\)

.

B

\(\displaystyle\int\dfrac {t×t}{(t-2)\,t}\;dt\)

C

\(\displaystyle\int \left (\dfrac {t}{t-2}×\dfrac {t-2}{t-2}\;\right)dt\)

D

\(\displaystyle\int \left (\dfrac {t}{t-2}×\dfrac {2}{2}\;\right)dt\)

Option A is Correct

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