Rational functions integration (HL) Recall from polynomial division that
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 ) This page is about integration of r ( x ) d ( x ) \displaystyle \frac{r(x)}{d(x)} d ( x ) r ( x ) d d d 2 2 2 r r r
We also assume that it is established that r r r d d d
The following cases depend on d ( x ) d(x) d ( x )
Contents Linear See indefinite integrals#linear over linear integration .
Quadratic with two real roots This is best solved using partial fractions . Certain problems can be solved with substitution, but it is not reliable, whereas method of partial fractions is reliable.
Example: Find
∫ x − 7 x 2 − 7 d x \displaystyle\int \frac{x - 7}{x^2 -7} \d x ∫ x 2 − 7 x − 7 d x
Denominator factors as ( x − 7 ) ( x + 7 ) (x - \sqrt 7)(x + \sqrt 7) ( x − 7 ) ( x + 7 )
x − 7 ( x − 7 ) ( x + 7 ) = A x − 7 + B x + 7 = A ( x + 7 ) + B ( x − 7 ) ( x − 7 ) ( x + 7 ) x − 7 = A ( x + 7 ) + B ( x − 7 ) \begin{align*} \frac{x - 7}{(x - \sqrt 7)(x + \sqrt 7)} &= \frac{A}{x - \sqrt 7} + \frac{B}{x + \sqrt 7} \\
&= \frac{A(x + \sqrt 7) + B(x - \sqrt 7)}{(x - \sqrt 7)(x + \sqrt 7)} \\
x - 7 &= A(x + \sqrt 7) + B(x - \sqrt 7)\end{align*} ( x − 7 ) ( x + 7 ) x − 7 x − 7 = x − 7 A + x + 7 B = ( x − 7 ) ( x + 7 ) A ( x + 7 ) + B ( x − 7 ) = A ( x + 7 ) + B ( x − 7 ) Evaluate at x = 7 x = \sqrt 7 x = 7
7 − 7 = 2 A 7 1 − 7 = 2 A 1 − 7 2 = A \begin{align*} \sqrt 7 - 7 &= 2A\sqrt 7 \\
1 - \sqrt7&= 2A \\
\frac{1 - \sqrt7}{2} &= A\end{align*} 7 − 7 1 − 7 2 1 − 7 = 2 A 7 = 2 A = A Evaluate at x = − 7 x = -\sqrt 7 x = − 7
− 7 − 7 = − 2 B 7 − 1 − 7 = 2 B 1 + 7 2 = B \begin{align*} -\sqrt 7 - 7 &= -2B\sqrt 7 \\ - 1 - \sqrt7 &= 2B \\\frac{1 + \sqrt7}{2} &= B\end{align*} − 7 − 7 − 1 − 7 2 1 + 7 = − 2 B 7 = 2 B = B I = ∫ ( 1 − 7 2 ( x − 7 ) + 1 + 7 2 ( x + 7 ) ) d x = 1 − 7 2 ln ∣ x − 7 ∣ + 1 + 7 2 ln ∣ x + 7 ∣ + C ■ \begin{align*} I &= \int \left(\frac{1 - \sqrt7}{2(x-\sqrt7)} + \frac{1 + \sqrt7}{2(x + \sqrt7)} \right) \d x \\
&= \frac{1 - \sqrt7}{2} \ln \lvert x-\sqrt7\rvert + \frac{1 + \sqrt7}{2} \ln \lvert x+\sqrt7\rvert + C \qed
\end{align*} I = ∫ ( 2 ( x − 7 ) 1 − 7 + 2 ( x + 7 ) 1 + 7 ) d x = 2 1 − 7 ln ∣ x − 7 ∣ + 2 1 + 7 ln ∣ x + 7 ∣ + C ■ You may also get 1 − 7 2 ln ∣ 2 x − 2 7 ∣ + 1 + 7 2 ln ∣ 2 x + 2 7 ∣ + C \displaystyle\frac{1 - \sqrt7}{2} \ln \lvert 2x-2\sqrt7\rvert + \frac{1 + \sqrt7}{2} \ln \lvert 2x+2\sqrt7\rvert + C 2 1 − 7 ln ∣ 2 x − 2 7 ∣ + 2 1 + 7 ln ∣ 2 x + 2 7 ∣ + C
They are equivalent as ln 2 a = ln 2 + ln a \ln 2a = \ln 2 + \ln a ln 2 a = ln 2 + ln a ln 2 \ln 2 ln 2 + C +C + C
Quadratic with no real roots This uses
∫ 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 and sometimes a substitution.
Example: Find
∫ 2 x + 1 2 x 2 + 4 x + 5 d x \displaystyle \int \frac{2x + 1}{2x^2 + 4x + 5} \d x ∫ 2 x 2 + 4 x + 5 2 x + 1 d x
We need to split this into
I = 1 2 ∫ 4 x + 4 − 2 2 x 2 + 4 x + 5 d x = 1 2 ∫ 4 x + 4 2 x 2 + 4 x + 5 d x − ∫ 1 2 x 2 + 4 x + 5 d x = 1 2 I 1 − I 2 \begin{align*}
I &= \frac12 \int \frac{4x + 4 - 2}{2x^2 + 4x + 5} \d x \\
&= \frac12 \int \frac{4x + 4}{2x^2 + 4x + 5} \d x - \int \frac{1}{2x^2 + 4x + 5} \d x \\
&= \frac12I_1 - I_2\end{align*} I = 2 1 ∫ 2 x 2 + 4 x + 5 4 x + 4 − 2 d x = 2 1 ∫ 2 x 2 + 4 x + 5 4 x + 4 d x − ∫ 2 x 2 + 4 x + 5 1 d x = 2 1 I 1 − I 2 Let u = 2 x 2 + 4 x + 5 > 0 , d u = 4 x + 4 u = 2x^2 + 4x + 5 > 0, \d u = 4x + 4 u = 2 x 2 + 4 x + 5 > 0 , d u = 4 x + 4
I 1 = ∫ d u u = ln u + C = ln ( 2 x 2 + 4 x + 5 ) + C \begin{align*} I_1 &= \int \frac{\d u}{u} \\
&= \ln u + C \\
&= \ln(2x^2 + 4x + 5) + C
\end{align*} I 1 = ∫ u d u = ln u + C = ln ( 2 x 2 + 4 x + 5 ) + C Now for I 2 I_2 I 2 complete the square .
2 x 2 + 4 x + 5 = 2 ( x 2 + 2 x ) + 5 = 2 ( x 2 + 2 x + 1 − 1 ) + 5 = 2 ( x + 1 ) 2 − 2 + 5 = ( 2 x + 2 ) 2 + 3 \begin{align*} 2x^2 + 4x + 5 &= 2(x^2 + 2x) + 5 \\
&= 2(x^2 + 2x + 1 - 1) + 5 \\
&= 2(x + 1)^2 - 2 + 5 \\
&= (\sqrt 2x + \sqrt2)^2 + 3
\end{align*} 2 x 2 + 4 x + 5 = 2 ( x 2 + 2 x ) + 5 = 2 ( x 2 + 2 x + 1 − 1 ) + 5 = 2 ( x + 1 ) 2 − 2 + 5 = ( 2 x + 2 ) 2 + 3 We are ready to use arctan \arctan arctan f ( a x + b ) f(ax + b) f ( a x + b )
I 2 = ∫ 1 ( 2 x + 2 ) 2 + ( 3 ) 2 d x = 1 2 3 arctan 2 x + 2 3 + C = 6 6 arctan 6 x + 6 3 + C \begin{align*} I_2 &= \int \frac{1}{(\sqrt 2x + \sqrt2)^2 + (\sqrt 3)^2} \d x \\
&= \frac{1}{\sqrt2\sqrt3} \arctan\frac{\sqrt2x+\sqrt2}{\sqrt3} + C \\
&= \frac{\sqrt6}{6} \arctan\frac{\sqrt6x+\sqrt6}{3} + C
\end{align*} I 2 = ∫ ( 2 x + 2 ) 2 + ( 3 ) 2 1 d x = 2 3 1 arctan 3 2 x + 2 + C = 6 6 arctan 3 6 x + 6 + C 1 2 I 2 − I 2 = 1 2 ln ( 2 x 2 + 4 x + 5 ) − 6 6 arctan 6 x + 6 3 + C ■ \begin{align*} \frac12 I_2 - I_2 = &\frac12 \ln(2x^2 + 4x + 5) \\ & - \frac{\sqrt6}{6} \arctan\frac{\sqrt6x+\sqrt6}{3} + C \qed\end{align*} 2 1 I 2 − I 2 = 2 1 ln ( 2 x 2 + 4 x + 5 ) − 6 6 arctan 3 6 x + 6 + C ■ Quadratic with a repeated root There is no point using partial fractions, since we can easily integrate ( a x + b ) − 2 (ax + b)^{-2} ( a x + b ) − 2
Substitutions can typically be bypassed as well.
Example: Find
∫ 3 x ( 3 x + 2 ) 2 d x \displaystyle \int \frac{3x}{(3x + 2)^2} \d x ∫ ( 3 x + 2 ) 2 3 x d x
We can to split this into reciprocal of a linear function and a reciprocal of a quadratic function.
I = ∫ ( 3 x + 2 ( 3 x + 2 ) 2 − 2 ( 3 x + 2 ) 2 ) d x = ∫ ( 1 3 x + 2 − 2 ( 3 x + 2 ) 2 ) d x = 1 3 ln ∣ 3 x + 2 ∣ + 2 3 ( 3 x + 2 ) + C ■ \begin{align*}
I &= \int \left(\frac{3x + 2}{(3x + 2)^2} - \frac{2}{(3x+2)^2}\right) \d x \\
&= \int \left(\frac{1}{3x + 2} - \frac{2}{(3x+2)^2}\right) \d x \\
&= \frac13\ln\lvert3x+2\rvert + \frac{2}{3(3x+2)} + C\qed\end{align*} I = ∫ ( ( 3 x + 2 ) 2 3 x + 2 − ( 3 x + 2 ) 2 2 ) d x = ∫ ( 3 x + 2 1 − ( 3 x + 2 ) 2 2 ) d x = 3 1 ln ∣ 3 x + 2 ∣ + 3 ( 3 x + 2 ) 2 + C ■ And both terms were divided by 3 3 3 f ( 3 x + 2 ) {f(3x + 2)} f ( 3 x + 2 )