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Integration of Rational Functions

If $f(x)$ and $g(x)$ are polynomials, then the integral of quotient of $f(x)$ and $g(x)$, which is $\displaystyle{\int \frac{f(x)}{g(x)}dx}$, can be obtained by the following method.

Partial Fractions
If $\deg f(x) > \deg g(x)$, then let $q(x)$ be the quotient of $f(x)$ and $g(x)$ and the remainder be $r(x)$. Thus

$\displaystyle \frac{f(x)}{g(x)} = q(x) + \frac{r(x)}{g(x)}, \ {\rm where}\ \deg r(x) < \deg g(x). $

Note that $q(x)$ is a polynomial. It is easy to integrate.

Factoring Denominator
Fundamental Theorem of Algebra states that every polynomial can be factored into the product of linear and quadratic polynomials.

Partial Fraction Decomposition
A partial fraction decomposition is the operation that consists in expressing the fraction as a sum of a polynomial and one or several fractions whose degree of numerator is one less than the degree of denominator.

Example 3..5   Decompose the following function by partial fraction expansion,

$\displaystyle \frac{3x}{(x-2)^3}$

NOTE The foctors of the denominator is $x-2, (x-2)^2, (x-2)^3$. Thus we need 3 partial fractions. Now inside of parenthesis, we have linear polynomial. Thus, the numerators must be constant.

$\displaystyle \frac{3x}{(x-2)^3} = \frac{A}{x - 2} + \frac{B}{(x - 2)^{2}} +\frac{C}{(x - 2)^{3}}\ensuremath{\ \blacksquare}$

Exercise 3..5   Decompose the following function by partial fraction expansion,

$\displaystyle \frac{3x}{(x^2 + 1)^3}$

SOLUTION The factors of denominator are, $x^2+1, (x^2+1)^2, (x^2+1)^3$. Thus we need three partial fractions. Now inside of parenthesis, we have quadratic polynomial. Thus, the numerators must be linear.

$\displaystyle \frac{3x}{(x^2+1)^3} = \frac{Ax+B}{x^2+1} + \frac{Cx+D}{(x^2+1)^{2}} +\frac{Ex+F}{(x^2+1)^{3}}\ensuremath{\ \blacksquare}$

Solving Partial Fraction by System of Linear Equations
Clear the denominator. Then we a polynomials in both sides of equation. Since the equation must be satisfied by all $x$, the coefficients of two polynomials of the same degree are equal. From this, we get a system of linear equations. Finally solve the system to get $A,B,C,D,\ldots$.

Example 3..6   Find the partial fraction of the following function.

$\displaystyle \frac{3x}{(x-2)^3}$

SOLUTION By Example3.5, $\frac{3x}{(x-2)^3} = \frac{A}{x - 2} + \frac{B}{(x - 2)^{2}} +\frac{C}{(x - 2)^{3}}$. Now clear the denominator and simplify to get
$\displaystyle 3x$ $\displaystyle =$ $\displaystyle A(x-2)^2 + B(x-2) + C = Ax^2 - 4Ax + 4A + Bx - 2B + C$  
  $\displaystyle =$ $\displaystyle Ax^2 + (-4A + B)x + (4A-2B+C).$  

Now the coefficients of two polynomials of the same degree are equal.

$\displaystyle A = 0, -4A + B = 3, 4A-2B+C = 0.$

Solving this to get $A = 0, \ B = 3, \ C = 6$,

$\displaystyle \frac{3x}{(x-2)^3} = \frac{3}{(x - 2)^{2}} +\frac{6}{(x - 2)^{3}}.$

Exercise 3..6   Find the partial fraction of the following function.

$\displaystyle \frac{3x}{(x^2 + 1)^3}$

SOLUTION $\frac{3x}{(x^2+1)^3} = \frac{Ax+B}{x^2+1} + \frac{Cx+D}{(x^2+1)^{2}} +\frac{Ex+F}{(x^2+1)^{3}}$. Clear the denominator,
$\displaystyle 3x$ $\displaystyle =$ $\displaystyle (Ax+B)(x^2 + 1)^2 + (Cx + D)(x^2 + 1) + Ex + F$  
  $\displaystyle =$ $\displaystyle Ax^5 + Bx^4 + 2Ax^3 + 2Bx^2 + Ax + B + Cx^3 + Dx^2 + Cx + D + Ex + F$  
  $\displaystyle =$ $\displaystyle Ax^5 + Bx^4 + (2A + C)x^3 + (2B+ D)x^2 + (A + C + E)x + (B + D + F).$  

Now $A = 0, B = 0, 2A + C = 0, 2B + D = 0, A + C+ E = 3, B + D+ F = 0$. Then $C = 0, D = 0$, Since $A + C +E = 3$, $E = 3$, $F = 0$. Therefore,

$\displaystyle \frac{3x}{(x^2+1)^3} = \frac{0}{x^2+1} + \frac{0}{(x^2+1)^{2}} +\frac{3x}{(x^2+1)^{3}}$

which is the same as the original fraction. If we let $t = x^2 + 1$, then we can solve this problem. $\ \blacksquare$

Partial Fraction of Rational Function
Every rational function $\displaystyle{\frac{f(x)}{g(x)}}$ can be decomposed by partial fraction to get a sum of the following functions..

$\displaystyle \frac{A}{(x - a)^{n}}, \ \ \frac{Bx+C}{\{(x-a)^{2}+b^{2}\}^{n}} $

If we let $t = x - a$. Then it is given in one of the following forms.

$\displaystyle \frac{1}{t^n},\ \ \frac{1}{(t^{2} + b^{2})^{n}}, \ \ \frac{t}{(t^{2} + b^{2})^{n}}. $

Put together, every integration of rational function can be solved by solving the following integrals.
Integration of Rational Function

Theorem 3..6  

$\displaystyle{1. \ \int \frac{dx}{(x-a)^{n}} = \left\{\begin{array}{ll} \frac{-... ...c & (n = 2,3,4,\ldots)\\ \log{\vert x-a\vert} + c & (n=1) \end{array}\right. }$

$\displaystyle{2. \ \int \frac{x}{(x^{2} +b^{2})^{n}}dx = \left\{\begin{array}{l... ...,3,4,\ldots)\\ \frac{1}{2}\log{( x^{2}+b^{2})} + c & (n=1) \end{array}\right.}$

$\displaystyle{3. \ {\rm Let}\ I_{n} = \int \frac{dx}{(x^{2} + A)^{n}}}$. Then the following recurrence relation holds.

$\displaystyle I_{n+1} = \frac{1}{2nA} \{\frac{x}{(x^{2}+A)^{n}} + (2n-1)I_{n}\} \ (n \geq 1)$

Proof 1. Let $t = x - a$. Then $dt = dx$ and

$\displaystyle \int \frac{dx}{(x-a)^n}$ $\displaystyle =$ $\displaystyle \int \frac{dt}{t^n} = \int t^{-n}dt$  

For $n=1$, we have
$\displaystyle \int t^{-1}dt$ $\displaystyle =$ $\displaystyle \int \frac{1}{t}dt = \log\vert t\vert + c = \log\vert x-a\vert + c.$  

For $n \neq 1$,
$\displaystyle \int t^{-n}dt$ $\displaystyle =$ $\displaystyle \frac{1}{-n+1}t^{-n+1} + c = \frac{1}{-n+1}(x-a)^{-n+1} + c.$  

2. Let $t = x^2 + b^2$. Then $dt = 2xdx$ and

$\displaystyle \int \frac{x dx}{(x^2 + b^2)^{n}}$ $\displaystyle =$ $\displaystyle \frac{1}{2}\int \frac{dt}{t^{n}} = \frac{1}{2}\int t^{-n} dt$  

For $n=1$, we have

$\displaystyle \int \frac{1}{2}t^{-1}dt$ $\displaystyle =$ $\displaystyle \frac{1}{2} \int \frac{1}{t}dt = \frac{1}{2}\log\vert t\vert + c = \frac{1}{2}\log(x^2 + b^2) + c.$  

For $n \neq 1$,
$\displaystyle \int \frac{1}{2}t^{-n}dt$ $\displaystyle =$ $\displaystyle \frac{1}{2} \big(\frac{1}{-n+1}\big)t^{-n+1} + c = \frac{1}{2} \big(\frac{1}{-n+1}\big)(x^2 + b^2)^{-n+1} + c.$  

3. Integration by parts, we have

$\left\{\begin{array}{lcl} f = \frac{1}{(x^2 + A)^n} && g' = 1\\ &\searrow&\\ ... ... \frac{-2nx(x^2+A)^{n-1}}{(x^2 + A)^{2n}} &\leftarrow& g = x \end{array}\right.$. Then

$\displaystyle I_{n} = \int \frac{dx}{(x^{2} + A)^{n}}$ $\displaystyle =$ $\displaystyle \frac{x}{(x^{2}+A)^{n}} + 2n\int\frac{x^{2}dx}{(x^{2}+A)^{n+1}}$  
  $\displaystyle =$ $\displaystyle \frac{x}{(x^{2}+A)^{n}} + 2n\int \frac{x^{2}+A-A}{(x^{2}+A)^{n+1}}dx$  
  $\displaystyle =$ $\displaystyle \frac{x}{(x^{2}+A)^{n}} + 2nI_{n} - 2nAI_{n+1} \ensuremath{\ \blacksquare}$  

Example 3..7   Find the following integral

$\displaystyle \int \frac{x^{4}}{x^{3} - 1} dx$

SOLUTION $\deg(x^4) > \deg(x^3 -1)$. Thus divide the numerator by the denominator.

$\displaystyle \frac{x^4}{x^3 -1} = \frac{x(x^3 -1) + x}{x^3 - 1} = x + \frac{x}{x^3 - 1}.$

Now factor the denominator.

$\displaystyle \frac{x^4}{x^3 -1} = x + \frac{x}{(x-1)(x^2 + x + 1)}.$

Then by partial fraction decomposition.
$\displaystyle \frac{x}{(x-1)(x^2 + x + 1)} = \frac{A}{x-1} + \frac{Bx + C}{x^2 + x + 1}.$      

Clear the denominator and simplify,

$\displaystyle x = A(x^2 + x + 1) + (x-1)(Bx + C) = (A + B)x^2 + (A - B +C)x + (A - C). $

Setting the coefficient of the same degree is equal.

$\displaystyle A+B = 0, A - B + C = 1, A- C = 0$

Solving $\displaystyle{A = \frac{1}{3}, \ B = -\frac{1}{3}, \ C = \frac{1}{3}}$. Thus
$\displaystyle \int \frac{x^{4}}{x^3 -1} dx$ $\displaystyle =$ $\displaystyle \int (x + \frac{1/3}{x -1} + \frac{-x/3 + 1/3}{x^2 + x + 1})dx$  
  $\displaystyle =$ $\displaystyle \frac{1}{2}x^{2} + \frac{1}{3}\log{\vert x-1\vert} - \frac{1}{3}[\int \frac{x-1}{x^2 + x+ 1} dx]$  


$\displaystyle \int \frac{x^{4}}{x^3 -1}$ $\displaystyle =$ $\displaystyle \frac{1}{2}x^{2} + \frac{1}{3}\log{\vert x-1\vert} - \frac{1}{3}[\int \frac{x-1}{(x +\frac{1}{2})^{2} + \frac{3}{4}} dx]$  

Let $\displaystyle{x + \frac{1}{2} = t}$. Then $dx = dt, x = t - \frac{1}{2}$.
$\displaystyle \int \frac{x-1}{(x +\frac{1}{2})^{2} + \frac{3}{4}} dx$ $\displaystyle =$ $\displaystyle \int \frac{t - 3/2}{t^2 + (\frac{\sqrt{3}}{2})^{2}} dt$  
  $\displaystyle =$ $\displaystyle \int \frac{t}{t^2 + (\frac{\sqrt{3}}{2})^{2}} dt - \frac{3}{2} \int \frac{1}{t^2 + (\frac{\sqrt{3}}{2})^{2}} dt$  
  $\displaystyle =$ $\displaystyle \frac{1}{2}\log{(t^2 + (\frac{\sqrt{3}}{2})^{2})} - \frac{3}{2}\frac{2}{\sqrt{3}}\tan^{-1}{\left(\frac{2t}{\sqrt{3}}\right)} + c$  
  $\displaystyle =$ $\displaystyle \frac{1}{2} \log{(x^2 + x+ 1)} - \sqrt{3}\tan^{-1}{\left(\frac{2x + 1}{\sqrt{3}}\right)} + c.$  

Exercise 3..7   Find the integral of the following.

$\displaystyle \int{\frac{dx}{(x^2 + 16)^2}}$

SOLUTION Let $I_n = \int \frac{1}{(x^2 + 16)^n}\: dx$. Then $\left\{\begin{array}{lcl} f = \frac{1}{(x^2 + 16)}, && g' = 1\\ &\searrow& \\ f' = -\frac{2x}{(x^2 + 16)^2}, &\leftarrow& g = x \end{array}\right.$

$\displaystyle I_{1}$ $\displaystyle =$ $\displaystyle \int{\frac{dx}{(x^2 + 16)}} = \frac{x}{(x^2 + 16)} + 2\int{\frac{x^2}{(x^2 + 16)^2}}dx$  
  $\displaystyle =$ $\displaystyle \frac{x}{(x^2 + 16)} + 2\int{\frac{x^2+16 -16}{(x^2 + 16)^2}}dx$  
  $\displaystyle =$ $\displaystyle \frac{x}{(x^2 + 16)} + 2\int{\frac{1}{(x^2 + 16)}}dx -32 \int{\frac{1}{(x^2 + 16)^2}}dx$  
  $\displaystyle =$ $\displaystyle \frac{x}{(x^2 + 16)} + 2I_{1} - 32I_{2}$  

Thus,

$\displaystyle I_{2} = \frac{1}{32}[\frac{x}{(x^2 + 16)} + I_{1}] = \frac{1}{128}[\frac{4x}{(x^2 + 16)} + \tan^{-1}(\frac{x}{4})] + c$

Exercise A


1.
Find the partial fraction expansion of the following functions?D

(a) $\displaystyle{f(x) = \frac{7}{(x-2)(x+5)} }$ (b) $\displaystyle{f(x) = \frac{x^2 + 1}{x(x^2 - 1)}}$ (c) $\displaystyle{f(x) = \frac{x^2 + 3}{x^2 - 3x + 2}}$ (d) $\displaystyle{f(x) = \frac{x^2}{(x - 1)^2(x + 1)}}$ (e) $\displaystyle{f(x) = \frac{x^{5}}{(x-2)^2}}$ (f) $\displaystyle{f(x) = \frac{x+1}{x(x^{2} + 1)}}$

Exercise B


1.
Work out the following integrals?D

(a) $\displaystyle{\int{\frac{7}{(x-2)(x+5)}}\ dx}$ (b) $\displaystyle{\int{\frac{x^2 + 1}{x(x^2 - 1)}}\ dx}$ (c) $\displaystyle{\int{\frac{x^2 + 3}{x^2 - 3x + 2}}\ dx}$

(d) $\displaystyle{\int{\frac{x^2}{(x - 1)^2(x + 1)}}\ dx}$ (e) $\displaystyle{\int{\frac{dx}{(x^2 + 16)^2}}}$ (f) $\displaystyle{\int{\frac{x^{5}}{(x-2)^2}}\ dx}$

(g) $\displaystyle{\int{\frac{x^5}{x^9 - 1}}\ dx \ \ (x^3 = t)}$ (h) $\displaystyle{\int{\frac{dx}{x(x^4 + 1)}} \ \ (x^4 = t)}$



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