Informative line

### 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 $$\dfrac{A}{x-a}+\dfrac{B}{x-b}+\dfrac{C}{(x-c)}$$

#### 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

#### 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

# 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

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

#### 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:

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

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

#### 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 $$Q(x)$$

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

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

#### 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

# 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.

#### 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