H2 Maths Formulas, Techniques & Graphs >> Calculus >> Integration techniques >>
Integrate Different Forms of Algebraic Fractions
Fractions in the form ${1 \over x^2 + a^2}$, ${1 \over \sqrt{a^2 - x^2} }$, ${1 \over x^2 - a^2}$, ${1 \over a^2 - x^2}$
| $$f(x)$$ | $$\int f(x) \phantom{.} dx $$ | Remarks |
|---|---|---|
| $$ {1 \over x^2 + a^2} $$ | $$ {1 \over a} \tan^{-1} \left(x \over a\right) $$ | On page 4 of MF26 |
| $$ {1 \over \sqrt{a^2 - x^2}} $$ | $$ \sin^{-1} \left(x \over a\right) $$ | On page 4 of MF26 |
| $$ {1 \over x^2 - a^2} $$ | $$ {1 \over 2a} \ln \left(x - a \over x + a\right) $$ | On page 4 of MF26 |
| $$ {1 \over a^2 - x^2} $$ | $$ {1 \over 2a} \ln \left(a + x \over a - x\right) $$ | On page 4 of MF26 |
To use the formulas directly, make sure the coefficient of $x^2$ is equals to 1. For example, \begin{align} \int {1 \over \sqrt{3 - 25x^2} } \phantom{.} dx & = \int { 1 \over \sqrt{25} \sqrt{ {3 \over 25} - x^2 } } \phantom{.} dx \\ & = {1 \over 5} \int { 1 \over \sqrt{ \left(\sqrt{3} \over 5\right)^2 - x^2 } } \phantom{.} dx \\ & = {1 \over 5} \sin^{-1} \left( x \over {\sqrt{3} \over 5} \right) \\ & = {1 \over 5} \sin^{-1} \left(5x \over \sqrt{3} \right) + C \end{align}
Fractions in the form ${1 \over \sqrt{ax^2 + bx+ c} }$
For this type of fraction, complete the square for the expression in the denominator and use the appropriate formula from MF26 (see previous section).
Example
$$ \int {1 \over \sqrt{2 - t^2 + 2t}} \phantom{.} dt $$
Complete the square for the expression in the denominator (without the square root): \begin{align} 2 - t^2 + 2t & = -t^2 + 2t + 2 \\ & = -(t^2 - 2t) + 2 \\ & = -[ (t - 1)^2 - (1)^2 ] + 2 \\ & = -(t - 1)^2 + 1 + 2 \\ & = -(t - 1)^2 + 3 \\ & = 3 - (t - 1)^2 \end{align}
Apply the result and use the appropriate formula from MF26: \begin{align} \int {1 \over \sqrt{2 - t^2 + 2t} } \phantom{.} dt & = \int {1 \over \sqrt{ 3 - (t - 1)^2} } \phantom{.} dt \\ & = \int {1 \over \sqrt{ (\sqrt{3})^2 - (t - 1)^2 } } \phantom{.} dt \\ & = \sin^{-1} \left( t - 1 \over \sqrt{3} \right) + C \end{align}
Fractions in the form ${1 \over ax^2 + bx+ c}$
There are two approaches for this type of fraction:
If the denominator is factorisable, do partial fractions
If the denominator is not factorisable, complete the square for the expression in the denominator and use the appropriate formula from MF26
Approach 1: Partial fractions
$$ \int {1 \over x^2 - 3x + 2} \phantom{.} dx $$
Since the denominator is factorisable, solve partial fraction (refer to page 2 of MF26 for the formulas): \begin{align} {1 \over x^2 - 3x + 2} & = {1 \over (x - 1)(x - 2)} \\ & = {A \over x - 1} + {B \over x - 2} \\ & = {A (x - 2) + B (x - 1) \over (x - 1)(x - 2)} \\ \\ 1 & = A(x - 2) + B(x - 1) \\ \\ \text{Let } & x = 2, \\ 1 & = B \\ \\ \text{Let } & x = 1, \\ 1 & = -A \\ -1 & = A \\ \\ {1 \over x^2 - 3x + 2} & = {-1 \over x - 1} + {1 \over x - 2} \end{align}
Now integrate: \begin{align} \int {1 \over x^2 - 3x + 2} \phantom{.} dx & = \int \left( {-1 \over x - 1} + {1 \over x - 2} \right) \phantom{.} dx \\ & = - \int {1 \over x - 1} \phantom{.} dx + \int {1 \over x - 2} \phantom{.} dx \\ & = - \ln |x - 1| + \ln |x - 2| + C \end{align}
Approach 2: Complete the square
$$ \int {1 \over x^2 - 3x + 5} \phantom{.} dx $$
Since the denominator is not factorisable, complete the square for the expression in the denominator: \begin{align} x^2 - 3x + 5 & = \left(x - {3 \over 2}\right)^2 - \left(3 \over 2\right)^2 + 5 \\ & = \left(x - {3 \over 2}\right)^2 + {11 \over 4} \end{align}
Now integrate, using the result and the appropriate formula from MF26: \begin{align} \int {1 \over x^2 - 3x + 5} \phantom{.} dx & = \int {1 \over \left(x - {3 \over 2}\right)^2 + {11 \over 4} } \phantom{.} dx \\ & = \int {1 \over \left(x - {3 \over 2}\right)^2 + \left( \sqrt{11} \over 2\right)^2 } \phantom{.} dx \\ & = {1 \over {\sqrt{11} \over 2} } \tan^{-1} \left(x - {3 \over 2} \over {\sqrt{11} \over 2} \right) \\ & = {2 \over \sqrt{11} } \tan^{-1} \left(2x - 3 \over \sqrt{11} \right) + C \end{align}
Proper fractions in the form ${f(x) \over g(x)}$
The formulas below can be used:
For $n \ne -1$, $$ \boxed{ \int f'(x) [f(x)]^n \phantom{.} dx = { [f(x)]^{n + 1} \over n + 1} + C } $$
For $n = -1$, $$ \boxed{ \int f'(x) [f(x)]^{-1} \phantom{.} dx = \int {f'(x) \over f(x)} \phantom{.} dx = \ln \left| f(x) \right| + C } $$
Example 1
\begin{align} \int {x - 1 \over x^2 - 2x - 7} \phantom{.} dx & = {1 \over 2} \int {2x - 2 \over x^2 - 2x - 7} \phantom{.} dx \\ & = {1 \over 2} \ln |x^2 - 2x - 7| + C \end{align}
Example 2
$$ \int { 2x \over \sqrt{5 - 4x - x^2} } \phantom{.} dx $$
Since $f(x) = 5 - 4x - x^2$ and $f'(x) = -4 - 2x$, manipulate the expression to obtain ${f'(x) \over f(x)}$: \begin{align} \int { 2x \over \sqrt{5 - 4x - x^2} } \phantom{.} dx & = - \int {-2x \over \sqrt{5 - 4x - x^2} } \phantom{.} dx \\ & = - \int \left( {- 2x - 4 \over \sqrt{5 - 4x - x^2} } + {4 \over \sqrt{5 - 4x - x^2} } \right) \phantom{.} dx \\ & = - \int {-2x - 4 \over \sqrt{5 - 4x - x^2} } \phantom{.} dx - \int {4 \over \sqrt{5 - 4x - x^2} } \phantom{.} dx \\ & = - \int (-2x - 4) (5 - 4x - x^2)^{-{1 \over 2}} \phantom{.} dx - 4 \int {1 \over \sqrt{5 - 4x - x^2} } \phantom{.} dx \end{align}
The first term is in the form of ${f'(x) \over f(x)}$. The second term can integrated by
- Complete the square for expression in denominator
- Use the relevant formula from MF26
\begin{align} 5 - 4x - x^2 & = -x^2 - 4x + 5 \\ & = -(x^2 + 4x) + 5 \\ & = - [ (x + 2)^2 - (2)^2 ] + 5 \\ & = -(x + 2)^2 + 4 + 5 \\ & = -(x + 2)^2 + 9 \\ & = 9 - (x + 2)^2 \\ & = 3^2 - (x + 2)^2 \\ \\ \int (-2x - 4) (5 - 4x - x^2)^{-{1 \over 2}} \phantom{.} dx - 4 \int {1 \over \sqrt{5 - 4x - x^2} } \phantom{.} dx & = - \left[ (5 - 4x - x^2)^{1 \over 2} \over {1 \over 2} \right] - 4 \int {1 \over \sqrt{3^2 - (x + 2)^2}} \phantom{.} dx \\ & = -2 \sqrt{5 - 4x - x^2} - 4 \sin^{-1} \left( x + 2 \over 3\right) + C \end{align}
Improper fractions in the form ${f(x) \over g(x)}$
Express the improper fraction as a proper fraction, then follow the same strategy as the previous section.
Example
\begin{align} \int {2x^2 \over x^2 + 3x + 5} \phantom{.} dx & = \int 2 + {-6x - 10 \over x^2 + 3x + 5} \phantom{.} dx \\ & = \int 2 \phantom{.} dx - \int {6x + 10 \over x^2 + 3x + 5} \phantom{.} dx \end{align}
For the proper fraction, $f(x) = x^2 + 3x + 5$ and $f'(x) = 2x + 3$: \begin{align} \phantom{\int {2x^2 \over x^2 + 3x + 5} . dx } & = 2x - \int \left( {6x + 9 \over x^2 + 3x + 5} + {1 \over x^2 + 3x + 5} \right) \phantom{.} dx \\ & = 2x - \int {6x + 9 \over x^2 + 3x + 5} \phantom{.} dx - \int {1 \over x^2 + 3x + 5} \phantom{.} dx \\ & = 2x - 3\int {2x + 3 \over x^2 + 3x + 5} \phantom{.} dx - \int {1 \over\left(x + {3 \over 2}\right)^2 - \left(3 \over 2\right)^2 + 5} \phantom{.} dx \\ & = 2x - 3\int {2x + 3 \over x^2 + 3x + 5} \phantom{.} dx - \int {1 \over \left(x + {3 \over 2}\right)^2 + {11 \over 4} } \phantom{.} dx \\ & = 2x - 3\int {2x + 3 \over x^2 + 3x + 5} \phantom{.} dx - \int {1 \over \left(x + {3 \over 2}\right)^2 + \left(\sqrt{11} \over 2\right)^2 } \phantom{.} dx \end{align}
The second term is in the form ${f'(x) \over f(x)}$ and the third term can be integrated by one of the formulas in MF26: \begin{align} \phantom{\int {2x^2 \over x^2 + 3x + 5} . dx } & = 2x - 3 \ln (x^2 + 3x + 5) - {1 \over {\sqrt{11} \over 2}} \tan^{-1} \left(x + {3 \over 2} \over {\sqrt{11} \over 2} \right) \\ & = 2x - 3 \ln (x^2 + 3x + 5) - {2 \over \sqrt{11}} \tan^{-1} \left(2x + 3 \over \sqrt{11}\right) + C \end{align}
Note $x^2 + 3x + 5 = \left(x + {3 \over 2}\right)^2 + \left(\sqrt{11} \over 2\right)^2 $ is always positive for all real values of $x$.