Partial fractions (HL) Contents Overview When dividing a polynomial by another of higher order, partial fractions can be helpful in integration .
p ( x ) d ( x ) = q ( x ) + r ( x ) d ( x ) \frac{p(x)}{d(x)} = q(x) + \frac{r(x)}{d(x)} d ( x ) p ( x ) = q ( x ) + d ( x ) r ( x ) Note: if the numerator is of the same or higher order than the denominator, first obtain the quotient q ( x ) q(x) q ( x ) polynomial division , prior to using partial fractions on r ( x ) d ( x ) \frac{r(x)}{d(x)} d ( x ) r ( x )
The first step to partial fractions is to factor the denominator, if it is not already factored.
The correct partial fraction decomposition is valid for all values of x x x x x x
Case 1: all factors are linear and distinct Example: Find the partial fraction decomposition of − 2 x + 13 2 x 2 − 5 x − 3 \displaystyle \frac{-2x + 13}{2x^2 -5x - 3} 2 x 2 − 5 x − 3 − 2 x + 13
The denominator factors as ( x − 3 ) ( 2 x + 1 ) (x - 3)(2x + 1) ( x − 3 ) ( 2 x + 1 )
− 2 x + 13 2 x 2 − 5 x − 3 = A x − 3 + B 2 x + 1 = A ( 2 x + 1 ) + B ( x − 3 ) ( x − 3 ) ( 2 x + 1 ) − 2 x + 13 = A ( 2 x + 1 ) + B ( x − 3 ) \begin{align*}\frac{-2x + 13}{2x^2 -5x - 3} &= \frac{A}{x - 3} + \frac{B}{2x + 1} \\ &=
\frac{A(2x + 1) + B(x - 3)}{(x-3)(2x + 1)} \\
-2x + 13 &= A(2x + 1) + B(x - 3)
\end{align*} 2 x 2 − 5 x − 3 − 2 x + 13 − 2 x + 13 = x − 3 A + 2 x + 1 B = ( x − 3 ) ( 2 x + 1 ) A ( 2 x + 1 ) + B ( x − 3 ) = A ( 2 x + 1 ) + B ( x − 3 ) When x = 3 x = 3 x = 3
− 2 ( 3 ) + 13 = 7 A A = 1 \begin{align*}-2(3)+13 &= 7A \\
A &= 1
\end{align*} − 2 ( 3 ) + 13 A = 7 A = 1 When x = − 1 2 x = -\frac{1}{2} x = − 2 1
14 = − 7 B 2 − 4 = B \begin{align*}14 &= \frac{-7B}{2} \\
-4 &= B
\end{align*} 14 − 4 = 2 − 7 B = B − 2 x + 13 2 x 2 − 5 x − 3 = 1 x − 3 − 4 2 x + 1 ■ \frac{-2x + 13}{2x^2 -5x - 3} = \frac{1}{x - 3} - \frac{4}{2x + 1} \qed 2 x 2 − 5 x − 3 − 2 x + 13 = x − 3 1 − 2 x + 1 4 ■ Case 2: factors are linear but not distinct This is not useful (in this course) since the original function can be easily integrated. So it is beyond the syllabus.
Example: Find the partial fraction decomposition of 4 x − 11 ( x − 2 ) 2 \displaystyle \frac{4x - 11}{(x-2)^2} ( x − 2 ) 2 4 x − 11
4 x − 11 ( x − 2 ) 2 = A x − 2 + B ( x − 2 ) 2 = A ( x − 2 ) + B ( x − 2 ) 2 4 x − 11 = A ( x − 2 ) + B \begin{align*}
\frac{4x - 11}{(x-2)^2} &= \frac{A}{x - 2} + \frac{B}{(x-2)^2} \\ &=
\frac{A(x-2) + B}{(x-2)^2} \\
4x - 11 &= A(x - 2) + B
\end{align*} ( x − 2 ) 2 4 x − 11 4 x − 11 = x − 2 A + ( x − 2 ) 2 B = ( x − 2 ) 2 A ( x − 2 ) + B = A ( x − 2 ) + B When x = 2 x = 2 x = 2
4 ( 2 ) − 11 = − 3 = B \begin{align*}4(2)-11 = -3 &= B
\end{align*} 4 ( 2 ) − 11 = − 3 = B B B B x = 3 x = 3 x = 3
1 = A + ( − 3 ) 4 = A \begin{align*} 1 &= A + (-3) \\ 4 &= A
\end{align*} 1 4 = A + ( − 3 ) = A 4 x − 11 ( x − 2 ) 2 = 4 x − 2 − 3 ( x − 2 ) 2 ■ \frac{4x - 11}{(x-2)^2}= \frac{4}{x - 2} - \frac{3}{(x-2)^2} \qed ( x − 2 ) 2 4 x − 11 = x − 2 4 − ( x − 2 ) 2 3 ■ Notes If integrating, and the denominator cannot be factored using real numbers, then complete the square .
∫ 1 x 2 + a 2 d x = 1 a arctan x a + C \int \frac{1}{x^2 + a^2}\d x = \frac 1a \arctan \frac xa + C ∫ x 2 + a 2 1 d x = a 1 arctan a x + C This technique is discussed in rational functions integration .