Math III Rational functions and partial fractions If the denominator factors, partial fractions turn a rational function into a sum of terms that integrate to logarithms.
Basic, Standard: Grade 12 · Term 2 / Advanced: Grade 12 · Exam prep
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Guide Examples Practice Related
The basic idea 1 ( x + 1 ) ( x + 3 ) \dfrac{1}{(x+1)(x+3)} ( x + 1 ) ( x + 3 ) 1 partial fractions ,
1 ( x + 1 ) ( x + 3 ) = 1 2 ( 1 x + 1 − 1 x + 3 ) \frac{1}{(x+1)(x+3)} = \frac{1}{2}\left(\frac{1}{x+1} - \frac{1}{x+3}\right) ( x + 1 ) ( x + 3 ) 1 = 2 1 ( x + 1 1 − x + 3 1 ) we can use ∫ d x x + a = log ∣ x + a ∣ + C \int \frac{dx}{x+a} = \log|x+a| + C ∫ x + a d x = log ∣ x + a ∣ + C
∫ d x ( x + 1 ) ( x + 3 ) = 1 2 ( log ∣ x + 1 ∣ − log ∣ x + 3 ∣ ) + C = 1 2 log ∣ x + 1 x + 3 ∣ + C \int \frac{dx}{(x+1)(x+3)} = \frac{1}{2}(\log|x+1| - \log|x+3|) + C = \frac{1}{2}\log\left|\frac{x+1}{x+3}\right| + C ∫ ( x + 1 ) ( x + 3 ) d x = 2 1 ( log ∣ x + 1∣ − log ∣ x + 3∣ ) + C = 2 1 log x + 3 x + 1 + C How to find partial fractions Set 3 x + 1 ( x − 1 ) ( x + 2 ) = A x − 1 + B x + 2 \dfrac{3x+1}{(x-1)(x+2)} = \dfrac{A}{x-1} + \dfrac{B}{x+2} ( x − 1 ) ( x + 2 ) 3 x + 1 = x − 1 A + x + 2 B ( x − 1 ) ( x + 2 ) (x-1)(x+2) ( x − 1 ) ( x + 2 )
3 x + 1 = A ( x + 2 ) + B ( x − 1 ) 3x + 1 = A(x+2) + B(x-1) 3 x + 1 = A ( x + 2 ) + B ( x − 1 ) This holds for all x x x x = 1 x = 1 x = 1 4 = 3 A 4 = 3A 4 = 3 A x = − 2 x = -2 x = − 2 − 5 = − 3 B -5 = -3B − 5 = − 3 B A = 4 3 A = \frac{4}{3} A = 3 4 B = 5 3 B = \frac{5}{3} B = 3 5
Tip
Substituting the values of x x x ( x + a ) 2 (x+a)^2 ( x + a ) 2 A x + a + B ( x + a ) 2 \dfrac{A}{x+a} + \dfrac{B}{(x+a)^2} x + a A + ( x + a ) 2 B
Divide first if the numerator has a high degree If the degree of the numerator is at least that of the denominator, divide first to get "polynomial + fraction".
x 2 + 3 x + 1 x − 1 = x + 4 + 5 x − 1 \frac{x^2 + 3x + 1}{x - 1} = x + 4 + \frac{5}{x-1} x − 1 x 2 + 3 x + 1 = x + 4 + x − 1 5 Then ∫ x 2 + 3 x + 1 x − 1 d x = 1 2 x 2 + 4 x + 5 log ∣ x − 1 ∣ + C \displaystyle\int \frac{x^2+3x+1}{x-1}\,dx = \frac{1}{2}x^2 + 4x + 5\log|x-1| + C ∫ x − 1 x 2 + 3 x + 1 d x = 2 1 x 2 + 4 x + 5 log ∣ x − 1∣ + C
When the numerator is the derivative of the denominator If the numerator is (a multiple of) the derivative of the denominator, as in 2 x + 1 x 2 + x + 1 \dfrac{2x+1}{x^2+x+1} x 2 + x + 1 2 x + 1 ∫ f ′ ( x ) f ( x ) d x = log ∣ f ( x ) ∣ + C \int \frac{f'(x)}{f(x)}\,dx = \log|f(x)| + C ∫ f ( x ) f ′ ( x ) d x = log ∣ f ( x ) ∣ + C
Common mistakes The coefficient : 1 ( x + a ) ( x + b ) = 1 x + a − 1 x + b \frac{1}{(x+a)(x+b)} = \frac{1}{x+a} - \frac{1}{x+b} ( x + a ) ( x + b ) 1 = x + a 1 − x + b 1 1 b − a \frac{1}{b-a} b − a 1 Skipping the division : partial fractions don't work while the numerator's degree is too high.Absolute values : write log ∣ x + a ∣ \log|x+a| log ∣ x + a ∣ Worked examples
Find the indefinite integral.
∫ x + 5 x − 2 d x \int \frac{x + 5}{x - 2}\,dx ∫ x − 2 x + 5 d x Check answer Hint Answer Solution + Worksheet ✓ Added
Hint When the degree of the numerator is at least that of the denominator, divide first to get "polynomial + fraction".
Answer ∫ x + 5 x − 2 d x = x + 7 log ∣ x − 2 ∣ + C \int \frac{x + 5}{x - 2}\,dx = x + 7\log|x - 2| + C ∫ x − 2 x + 5 d x = x + 7 log ∣ x − 2∣ + C (C C C
Solution Dividing the numerator by the denominator, x + 5 = ( x − 2 ) + 7 x + 5 = (x - 2) + 7 x + 5 = ( x − 2 ) + 7
x + 5 x − 2 = 1 + 7 x − 2 \begin{aligned}\frac{x + 5}{x - 2} &= 1 + \frac{7}{x - 2}\end{aligned} x − 2 x + 5 = 1 + x − 2 7 Therefore
∫ x + 5 x − 2 d x = ∫ ( 1 + 7 x − 2 ) d x = x + 7 log ∣ x − 2 ∣ + C \begin{aligned}\int \frac{x + 5}{x - 2}\,dx &= \int \left(1 + \frac{7}{x - 2}\right)\,dx \\ &= x + 7\log|x - 2| + C\end{aligned} ∫ x − 2 x + 5 d x = ∫ ( 1 + x − 2 7 ) d x = x + 7 log ∣ x − 2∣ + C
Find the indefinite integral.
∫ 2 x 2 − x − 6 d x \int \frac{2}{x^{2} - x - 6}\,dx ∫ x 2 − x − 6 2 d x Check answer Hint Answer Solution + Worksheet ✓ Added
Hint Factor the denominator, then use partial fractions.
Answer ∫ 2 x 2 − x − 6 d x = 2 5 log ∣ x − 3 x + 2 ∣ + C \int \frac{2}{x^{2} - x - 6}\,dx = \frac{2}{5}\log\left|\frac{x - 3}{x + 2}\right| + C ∫ x 2 − x − 6 2 d x = 5 2 log x + 2 x − 3 + C (C C C
Solution Factoring the denominator, x 2 − x − 6 = ( x − 3 ) ( x + 2 ) x^{2} - x - 6 = (x - 3)(x + 2) x 2 − x − 6 = ( x − 3 ) ( x + 2 )
2 x 2 − x − 6 = 2 5 ( 1 x − 3 − 1 x + 2 ) \begin{aligned}\frac{2}{x^{2} - x - 6} &= \frac{2}{5}\left(\frac{1}{x - 3} - \frac{1}{x + 2}\right)\end{aligned} x 2 − x − 6 2 = 5 2 ( x − 3 1 − x + 2 1 ) Therefore
∫ 2 x 2 − x − 6 d x = ∫ 2 5 ( 1 x − 3 − 1 x + 2 ) d x = 2 5 log ∣ x − 3 x + 2 ∣ + C \begin{aligned}\int \frac{2}{x^{2} - x - 6}\,dx &= \int \frac{2}{5}\left(\frac{1}{x - 3} - \frac{1}{x + 2}\right)\,dx \\ &= \frac{2}{5}\log\left|\frac{x - 3}{x + 2}\right| + C\end{aligned} ∫ x 2 − x − 6 2 d x = ∫ 5 2 ( x − 3 1 − x + 2 1 ) d x = 5 2 log x + 2 x − 3 + C
Find the indefinite integral.
∫ x 3 x 2 + 4 d x \int \frac{x^{3}}{x^{2} + 4}\,dx ∫ x 2 + 4 x 3 d x Check answer Hint Answer Solution + Worksheet ✓ Added
Hint After dividing to lower the degree, a term of the form f ′ ( x ) f ( x ) \frac{f'(x)}{f(x)} f ( x ) f ′ ( x )
Answer ∫ x 3 x 2 + 4 d x = 1 2 x 2 − 2 log ( x 2 + 4 ) + C \int \frac{x^{3}}{x^{2} + 4}\,dx = \frac{1}{2}x^{2} - 2\log(x^{2} + 4) + C ∫ x 2 + 4 x 3 d x = 2 1 x 2 − 2 log ( x 2 + 4 ) + C (C C C
Solution Since x 3 = x ( x 2 + 4 ) − 4 x x^3 = x(x^{2} + 4) - 4x x 3 = x ( x 2 + 4 ) − 4 x
x 3 x 2 + 4 = x − 4 x x 2 + 4 \begin{aligned}\frac{x^{3}}{x^{2} + 4} &= x - \frac{4x}{x^{2} + 4}\end{aligned} x 2 + 4 x 3 = x − x 2 + 4 4 x Since ( x 2 + 4 ) ′ = 2 x (x^{2} + 4)' = 2x ( x 2 + 4 ) ′ = 2 x ∫ x x 2 + 4 d x = 1 2 log ( x 2 + 4 ) + C \displaystyle\int \frac{x}{x^{2} + 4}\,dx = \frac{1}{2}\log(x^{2} + 4) + C ∫ x 2 + 4 x d x = 2 1 log ( x 2 + 4 ) + C
∫ x 3 x 2 + 4 d x = ∫ ( x − 4 x x 2 + 4 ) d x = 1 2 x 2 − 2 log ( x 2 + 4 ) + C \begin{aligned}\int \frac{x^{3}}{x^{2} + 4}\,dx &= \int \left(x - \frac{4x}{x^{2} + 4}\right)\,dx \\ &= \frac{1}{2}x^{2} - 2\log(x^{2} + 4) + C\end{aligned} ∫ x 2 + 4 x 3 d x = ∫ ( x − x 2 + 4 4 x ) d x = 2 1 x 2 − 2 log ( x 2 + 4 ) + C Practice problems
Find the indefinite integral.
∫ 2 x − 3 x − 2 d x \int \frac{2x - 3}{x - 2}\,dx ∫ x − 2 2 x − 3 d x Check answer Hint Answer Solution + Worksheet ✓ Added
Hint When the degree of the numerator is at least that of the denominator, divide first to get "polynomial + fraction".
Answer ∫ 2 x − 3 x − 2 d x = 2 x + log ∣ x − 2 ∣ + C \int \frac{2x - 3}{x - 2}\,dx = 2x + \log|x - 2| + C ∫ x − 2 2 x − 3 d x = 2 x + log ∣ x − 2∣ + C (C C C
Solution Dividing the numerator by the denominator, 2 x − 3 = 2 ( x − 2 ) + 1 2x - 3 = 2(x - 2) + 1 2 x − 3 = 2 ( x − 2 ) + 1
2 x − 3 x − 2 = 2 + 1 x − 2 \begin{aligned}\frac{2x - 3}{x - 2} &= 2 + \frac{1}{x - 2}\end{aligned} x − 2 2 x − 3 = 2 + x − 2 1 Therefore
∫ 2 x − 3 x − 2 d x = ∫ ( 2 + 1 x − 2 ) d x = 2 x + log ∣ x − 2 ∣ + C \begin{aligned}\int \frac{2x - 3}{x - 2}\,dx &= \int \left(2 + \frac{1}{x - 2}\right)\,dx \\ &= 2x + \log|x - 2| + C\end{aligned} ∫ x − 2 2 x − 3 d x = ∫ ( 2 + x − 2 1 ) d x = 2 x + log ∣ x − 2∣ + C
Find the indefinite integral.
∫ 1 x 2 − 9 d x \int \frac{1}{x^{2} - 9}\,dx ∫ x 2 − 9 1 d x Check answer Hint Answer Solution + Worksheet ✓ Added
Hint Factor the denominator as ( x − a ) ( x + a ) (x-a)(x+a) ( x − a ) ( x + a )
Answer ∫ 1 x 2 − 9 d x = 1 6 log ∣ x − 3 x + 3 ∣ + C \int \frac{1}{x^{2} - 9}\,dx = \frac{1}{6}\log\left|\frac{x - 3}{x + 3}\right| + C ∫ x 2 − 9 1 d x = 6 1 log x + 3 x − 3 + C (C C C
Solution Factoring x 2 − 9 = ( x − 3 ) ( x + 3 ) x^{2} - 9 = (x - 3)(x + 3) x 2 − 9 = ( x − 3 ) ( x + 3 )
1 x 2 − 9 = 1 6 ( 1 x − 3 − 1 x + 3 ) \begin{aligned}\frac{1}{x^{2} - 9} &= \frac{1}{6}\left(\frac{1}{x - 3} - \frac{1}{x + 3}\right)\end{aligned} x 2 − 9 1 = 6 1 ( x − 3 1 − x + 3 1 ) Therefore
∫ 1 x 2 − 9 d x = ∫ 1 6 ( 1 x − 3 − 1 x + 3 ) d x = 1 6 log ∣ x − 3 x + 3 ∣ + C \begin{aligned}\int \frac{1}{x^{2} - 9}\,dx &= \int \frac{1}{6}\left(\frac{1}{x - 3} - \frac{1}{x + 3}\right)\,dx \\ &= \frac{1}{6}\log\left|\frac{x - 3}{x + 3}\right| + C\end{aligned} ∫ x 2 − 9 1 d x = ∫ 6 1 ( x − 3 1 − x + 3 1 ) d x = 6 1 log x + 3 x − 3 + C
Find the indefinite integral.
∫ x 3 x 2 − 4 d x \int \frac{x^{3}}{x^{2} - 4}\,dx ∫ x 2 − 4 x 3 d x Check answer Hint Answer Solution + Worksheet ✓ Added
Hint After dividing to lower the degree, a term of the form f ′ ( x ) f ( x ) \frac{f'(x)}{f(x)} f ( x ) f ′ ( x )
Answer ∫ x 3 x 2 − 4 d x = 1 2 x 2 + 2 log ∣ x 2 − 4 ∣ + C \int \frac{x^{3}}{x^{2} - 4}\,dx = \frac{1}{2}x^{2} + 2\log|x^{2} - 4| + C ∫ x 2 − 4 x 3 d x = 2 1 x 2 + 2 log ∣ x 2 − 4∣ + C (C C C
Solution Since x 3 = x ( x 2 − 4 ) + 4 x x^3 = x(x^{2} - 4) + 4x x 3 = x ( x 2 − 4 ) + 4 x
x 3 x 2 − 4 = x + 4 x x 2 − 4 \begin{aligned}\frac{x^{3}}{x^{2} - 4} &= x + \frac{4x}{x^{2} - 4}\end{aligned} x 2 − 4 x 3 = x + x 2 − 4 4 x Since ( x 2 − 4 ) ′ = 2 x (x^{2} - 4)' = 2x ( x 2 − 4 ) ′ = 2 x ∫ x x 2 − 4 d x = 1 2 log ∣ x 2 − 4 ∣ + C \displaystyle\int \frac{x}{x^{2} - 4}\,dx = \frac{1}{2}\log|x^{2} - 4| + C ∫ x 2 − 4 x d x = 2 1 log ∣ x 2 − 4∣ + C
∫ x 3 x 2 − 4 d x = ∫ ( x + 4 x x 2 − 4 ) d x = 1 2 x 2 + 2 log ∣ x 2 − 4 ∣ + C \begin{aligned}\int \frac{x^{3}}{x^{2} - 4}\,dx &= \int \left(x + \frac{4x}{x^{2} - 4}\right)\,dx \\ &= \frac{1}{2}x^{2} + 2\log|x^{2} - 4| + C\end{aligned} ∫ x 2 − 4 x 3 d x = ∫ ( x + x 2 − 4 4 x ) d x = 2 1 x 2 + 2 log ∣ x 2 − 4∣ + C