Learn definition of partial fractions with examples and cases. Practice partial fraction decomposition of the function & integration substitution.
For example:
\(\dfrac {1}{x+3} - \dfrac{2}{x+1}=\dfrac{(x+1)-2(x+3)}{(x+3)(x+1)}\)
\(\implies \)\(\dfrac {1}{x+3} - \dfrac{2}{x+1}=\dfrac{(-x-5)}{x^2+4x+3}\)
\(\displaystyle\int \dfrac{-x-5}{x^2+4x+3}dx=\int\dfrac{1}{x+3}dx-\int\dfrac{2}{x+1}dx\)
\(=\ell n |x+3|-2\ell n|x+1|+C\)
Here, we are discussing the case,
\(\dfrac{P(x)}{Q(x)}=\dfrac{A_1}{a_1x+b_1}+\dfrac{A_2}{a_2x+b_2}+......+\dfrac{A_n}{a_nx+b_n}\)
where \(A_1,A_2,A_3....A_n\) are constants and can be determined.
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 \(\dfrac{A}{x-a}+\dfrac{B}{x-b}+\dfrac{C}{(x-c)}\)
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 \(f(x)=\dfrac{A_1}{(x+7)}+\dfrac{A_2}{(2x+3)}\)
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 \(f(x)=\dfrac{A}{x-12}+\dfrac{B}{3x+5}+\dfrac{C}{x}\)
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 \(f(x)=\dfrac{Ax}{5x+1}+\dfrac{C}{x+1}\)
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 \(\dfrac{A_1x+B_1}{(ax^2+bx+c)}+\dfrac{A_2x+B_2}{(ax^2+bx+c)^2}\)
\(\dfrac{P(x)}{(ax^2+bx+c)^n}\Rightarrow\;\dfrac{A_1x+B_1}{(ax^2+bx+c)}+\dfrac{A_2x+B_2}{(ax^2+bx+c)^2}+\,.....\,\dfrac{A_nx+B_n}{(ax^2+bx+c)^n}\)
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
\(f(x)=\dfrac{A_1x+B_1}{(x^2+3x+1)}+\dfrac{A_2x+B_2}{(x^2+3x+1)^2}+\dfrac{A_3}{x+3}+\dfrac{A_4}{(x+3)^2}\)
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 \(f(x)=\dfrac{A}{(x^2+x+1)^2}-\dfrac{B}{x+1}\)
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:
\(1.\;\dfrac{P(x)}{Q(x)}=\dfrac{x^3-4x-10}{x^2-x-6}=x+1+\dfrac{3x-4}{x^2-x-6}\)
\(2.\;\dfrac{P(x)}{Q(x)}=\dfrac{2x^3-11x^2-2x+2}{2x^2+x-1}=(x-6)+\dfrac{5x-4}{2x^2+x-1}\)
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 \((9\,x-1) +\dfrac{1}{(2x+1)} +\dfrac{1}{(3x+1)}\)
\(1.\;\displaystyle\int\dfrac {t^2}{t^2-2}\;dt\)
It seems quite difficult.
\(\displaystyle\int\dfrac {t^2-2+2}{t^2-2}\;dt\)
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.
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 \(\displaystyle\int \left (\dfrac {t}{t-2}×\dfrac {2}{2}\;\right)dt\)