]> Special Methods

### Chapter 2: Integration; Section 1: Indefinite Integrals; page 3

Previous topic: page 2 Methods

Next topic: page 4 Simple Differential Equations

# Special Methods

## Rational Functions

A rational function is defined as the ratio of two polynomials (1.2.2.6):

 (2.1.3.1)

Any rational function can be integrated and we are going to learn how.

By the help of (2.1.2.2), one can intregrate the function:

 ${F}_{1}\left(x\right)=\frac{{a}_{0}}{{b}_{0}x+{b}_{1}}\text{ }\text{ }\left\{\begin{array}{l}m=0\\ n=1\end{array}\right\}$ (2.1.3.2)

and by the help of (2.1.2.3-4) - the function:

 ${F}_{2}\left(x\right)=\frac{{a}_{0}x+{a}_{1}}{{b}_{0}{x}^{2}+{b}_{1}x+{b}_{2}}\text{ }\text{ }\left\{\begin{array}{l}m=1\\ n=2\end{array}\right\}$ (2.1.3.3)

With the aid of integration by parts, one can extend the integration of these functions to the functions:

 $\begin{array}{l}{F}_{1p}\left(x\right)=\frac{{a}_{0}}{{\left({b}_{0}x+{b}_{1}\right)}^{p}}\\ {F}_{2q}\left(x\right)=\frac{{a}_{0}x+{a}_{1}}{{\left({b}_{0}{x}^{2}+{b}_{1}x+{b}_{2}\right)}^{q}}\end{array}\right\}\text{ }p,\text{ }q=\text{natural}\text{\hspace{0.28em}}\text{numbers}$ (2.1.3.4)

It will be shown that the integrals of (2.1.3.4) are enough, for integrating any rational function (2.1.3.1) with  m<n . We'll see first how to reduce a rational function with  $m\ge n$  to that form.

When we have a rational number expressed as a ratio of integers with a numerator, whose absolute value is bigger than that of the denominator, we can write it as an integer with the addition of a partial fraction (a number whose absolute value is smaller than one). This is done by division. A number expressed in decimal form is similar to a polynomial with  x=10 . Similarly to the ratio of integers, a division of polynomials can be effected so, as to obtain a polynomial with the addition of a rational function in the form of a partial fraction, meaning  m<n  at (2.1.3.1). One should notice the different defininitions of partial fraction in the cases of rational functions and rational numbers.

The technique of dividing two polynomials is similar but not identical to that of integers. For some time in the 20th century, people used to divide numbers by hand. Although we learned in school how this is done, it is almost forgotten, since it is much easier to use a calculator. This is why we need a refresher for numbers and a demonstration of the division of polynomials presented in Fig. Division.

The examples used for this demonstration are a division of two numbers

 $\frac{4925}{23}=214+\frac{3}{23}$ (2.1.3.5)

and a division between two polynomials

 $\frac{4{x}^{3}+9{x}^{2}+2x+5}{2x+3}=2{x}^{2}+\frac{3}{2}x-\frac{5}{4}+\frac{35}{4\left(2x+3\right)}$ (2.1.3.6)

The polynomials from the ratio (2.1.3.6) were chosen such that the substitution  x = 10  yields the same ratio as the numbers of (2.1.3.5), therefore it is easy to compare the results. Since the procedures are different, the results are also different. Using the result of the division of the polynomials, the integration of the rational function (2.1.3.6) becomes trivial:

 $\begin{array}{l}\int \frac{4{x}^{3}+9{x}^{2}+2x+5}{2x+3}\text{​}\text{d}x=\\ =\frac{2{x}^{3}}{3}+\frac{3{x}^{2}}{4}-\frac{5x}{4}+\frac{35}{8}\mathrm{ln}|2x+3|+C\end{array}\right\}$ (2.1.3.7)

where the sibstitution  u = (2x+3)  was used.

In the general case after the division of polynomials, the obtained partial fraction of polynomials should be of the form (2.1.3.1) restricted to m<n, however not necessarily of the the integrable form (2.1.3.4). Fortunately any partial fraction of polynomials can be expressed as a sum of the integrable forms (2.1.3.4).

It can be proven that any polynomial can be factorized as a product of polynomials of the first and the second order. The natural place to prove this lemma is within the context of complex numbers, and we are not presenting the proof here. Without entering to the general case, we are going to use the factorization only for some simple cases.

We can use the following example of a polynomial to illustrate factorization

 $\begin{array}{l}{P}_{1}\left(x\right)=x-1\text{ }\text{is}\text{\hspace{0.28em}}\text{a}\text{\hspace{0.28em}}\text{factor}\text{\hspace{0.28em}}\text{of}\text{ }{P}_{3}\left(x\right)={x}^{3}+{x}^{2}-2\\ \frac{{P}_{3}\left(x\right)}{{P}_{1}\left(x\right)}=\frac{{x}^{3}+{x}^{2}-2}{x-1}={x}^{2}+2x+2={P}_{2}\left(x\right)\\ {P}_{3}\left(x\right)={P}_{1}\left(x\right){P}_{2}\left(x\right)=\left(x-1\right)\left({x}^{2}+2x+2\right)\end{array}\right\}$ (2.1.3.8)

where a division of two polynomials was used.

In order to integrate a rational function expressed as a partial ratio, one has first to factorize the denominator and then write the function as a sum of integrable functions according to (2.1.3.4). If any of the factors do not appear more than once, the integrable functions will be of the forms (2.1.3.2) and (2.1.3.3).

As an example we'll use a rational function with the denominator P3(x) from the previous example (2.1.3.8).

 $\begin{array}{l}F\left(x\right)=\frac{5{x}^{2}-x+1}{{x}^{3}+{x}^{2}-2}=\frac{5{x}^{2}-x+1}{\left({x}^{2}+2x+2\right)\left(x-1\right)}=\\ =\frac{Ax+B}{{x}^{2}+2x+2}+\frac{C}{x-1}=\\ =\frac{\left(A+C\right){x}^{2}+\left(-A+B+2C\right)x+\left(-B+2C\right)}{{x}^{3}+{x}^{2}-2}\\ A,\text{ }B,\text{ }C\text{ }\text{are}\text{\hspace{0.28em}}\text{obtained}\text{\hspace{0.28em}}\text{by}\text{\hspace{0.28em}}\text{comparison :}\\ \left\{\begin{array}{l}C+A=5\\ 2C+B-A=-1\\ 2C-B=1\end{array}\right\}⇒\left\{\begin{array}{l}A=4\\ B=1\\ C=1\end{array}\right\}\\ F\left(x\right)=\frac{5{x}^{2}-x+1}{{x}^{3}+{x}^{2}-2}=\frac{4x+1}{{x}^{2}+2x+2}+\frac{1}{x-1}\end{array}\right\}$ (2.1.3.9)

On the other hand if the factorization of the denominator yields a factor that appears not once but n times, one has to add in the summation terms of the type (2.1.3.4) with powers p = 1,2,....n. Here is an example for completeness.

 $\begin{array}{l}F\left(x\right)=\frac{{a}_{0}{x}^{3}+{a}_{1}{x}^{2}+{a}_{2}x+{a}_{3}}{{x}^{4}-{x}^{2}-2x+2}=\\ =\frac{{a}_{0}{x}^{3}+{a}_{1}{x}^{2}+{a}_{2}x+{a}_{3}}{\left({x}^{2}+2x+2\right){\left(x-1\right)}^{2}}=\\ =\frac{Ax+B}{{x}^{2}+2x+2}+\frac{{C}_{1}}{x-1}+\frac{{C}_{2}}{{\left(x-1\right)}^{2}}\end{array}\right\}$ (2.1.3.10)

The constants $A,B,{C}_{1},{C}_{2}$ can be expressed in terms of ${a}_{0},{a}_{1},{a}_{2},{a}_{3}$ by comparison, as in the case of (2.1.3.9).

## Trigonometric Substitution

We already saw a parametric representation of the trigonometric function (1.2.3.9-11) which are reproduced here as (2.1.3.11):

 $\begin{array}{l}t=\mathrm{tan}\frac{\theta }{2}\\ \left\{\begin{array}{l}\mathrm{sin}\theta =\frac{2t}{1+{t}^{2}}\\ \mathrm{cos}\theta =\frac{1-{t}^{2}}{1+{t}^{2}}\\ \mathrm{tan}\theta =\frac{2t}{1-{t}^{2}}\end{array}\end{array}\right\}$ (2.1.3.11)

from which one obtains the differential relation

 $\text{d}\theta =\frac{2\text{d}t}{1+{t}^{2}}$ (2.1.3.12)

The substitution (2.1.3.11-12) applied on any algebraic expression of the trigonometric functions transforms it in a rational function, which is always integrable. Even if this substitution always works, it does not mean that the integral obtained is simpler, than by using a different method.

The following integral was solved by simple substitution at (2.1.2.5-6):

$\int \mathrm{sin}x\mathrm{cos}x\text{​}\text{d}x=\frac{{\mathrm{sin}}^{2}x}{2}+C$

The implementation of the substitution (2.1.3.11-12) for this example gives a soluble but not a simple integral:

 $\begin{array}{l}\int \mathrm{sin}x\mathrm{cos}x\text{​}\text{d}x=4\int \frac{t\left(1-{t}^{2}\right)\text{d}t}{{\left(1+{t}^{2}\right)}^{3}}=\\ =2\int \frac{\left(1-{t}^{2}\right)\text{d}\left(1+{t}^{2}\right)}{{\left(1+{t}^{2}\right)}^{3}}=\\ =2\int \frac{\left[2-\left(1+{t}^{2}\right)\right]\text{d}\left(1+{t}^{2}\right)}{{\left(1+{t}^{2}\right)}^{3}}=\\ =4\int \frac{\text{d}\left(1+{t}^{2}\right)}{{\left(1+{t}^{2}\right)}^{3}}-2\int \frac{\text{d}\left(1+{t}^{2}\right)}{{\left(1+{t}^{2}\right)}^{2}}=\\ =-\frac{2}{{\left(1+{t}^{2}\right)}^{2}}+\frac{2}{\left(1+{t}^{2}\right)}+C=\\ =\frac{2{t}^{2}}{{\left(1+{t}^{2}\right)}^{2}}+C=\frac{{\mathrm{sin}}^{2}x}{2}+C\end{array}\right\}$ (2.1.3.13)

On the other hand the substitution (2.1.3.11-12) gives a simple solution in the following example:

 $\int \frac{\text{d}x}{\mathrm{sin}x}=\int \frac{\text{d}t}{t}=\mathrm{ln}|t|+C=\mathrm{ln}|\mathrm{tan}\frac{x}{2}|+C$ (2.1.3.14)

## Table of Integrals

The solution of indefinite integrals can some times be tedious, time-consuming and even irritating. Fortunately there are tables of integrals, that include practically all the solvable integrals that can be expressed by elementary functions, and are needed in science and engineering. Such tables are part of the different mathematical handbooks. A good handbook contains much more than the table of integrals and is a requirement for studying science and engineering.

We'll take one example in order to see how it works. We want to solve the following integral:

 $\int \sqrt{\frac{2x-3}{2x+1}}\text{d}x=?$ (2.1.3.15)

We look in the table for integrals involving  $\sqrt{{a}_{0}x+{a}_{1}}\text{ }\text{and}\text{ }\sqrt{{b}_{0}x+{b}_{1}}$  and we find:

 $\begin{array}{l}\int \sqrt{\frac{{a}_{0}x+{a}_{1}}{{b}_{0}x+{b}_{1}}}\text{d}x=\\ =\left\{\begin{array}{l}\frac{\sqrt{\left({a}_{0}x+a\right)\left({b}_{0}x+{b}_{1}\right)}}{{b}_{0}}+\\ +\frac{{a}_{1}{b}_{0}-{a}_{0}{b}_{1}}{2{b}_{0}}\int \frac{\text{d}x}{\sqrt{\left({a}_{0}x+a\right)\left({b}_{0}x+{b}_{1}\right)}}\end{array}\end{array}\right\}$ (2.1.3.16)

There is an integral left in (2.1.3.16) that we have to find in the tables. Indeed

 $\begin{array}{l}\int \frac{\text{d}x}{\sqrt{\left({a}_{0}x+{a}_{1}\right)\left({b}_{0}x+{b}_{1}\right)}}=\\ =\text{\hspace{0.17em}}\text{or}\text{\hspace{0.17em}}\left\{\begin{array}{l}\frac{2}{\sqrt{{a}_{0}{b}_{0}}}\mathrm{ln}\left[\sqrt{{a}_{0}\left({b}_{0}x+{b}_{1}\right)}+\sqrt{{b}_{0}\left({a}_{0}x+{a}_{1}\right)}\right]\\ \frac{2}{\sqrt{-{a}_{0}{b}_{0}}}\text{atan}\sqrt{\frac{-{a}_{0}\left({b}_{0}x+{b}_{1}\right)}{{b}_{0}\left({a}_{0}x+{a}_{1}\right)}}\end{array}\end{array}\right\}$ (2.1.3.17)

where the choice between the upper and lower "or" branch depends on the sign of the  a0b0  product.

After applying (2.1.3.16-17) at (2.1.3.15) one obtains

 $\begin{array}{l}\int \sqrt{\frac{2x-3}{2x+1}}\text{d}x=\\ =\left\{\begin{array}{l}\frac{\sqrt{\left(2x-3\right)\left(2x+1\right)}}{2}+\\ +\frac{-6-2}{4}\int \frac{\text{d}x}{\sqrt{\left(2x-3\right)\left(2x+1\right)}}=\end{array}\\ =\left\{\begin{array}{l}\frac{\sqrt{\left(2x-3\right)\left(2x+1\right)}}{2}-\\ -2\mathrm{ln}\left[\sqrt{2\left(2x+1\right)}+\sqrt{2\left(2x-3\right)}\right]+\\ C\end{array}\end{array}\right\}$ (2.1.3.18)

## Exercises

Exercise 1. Calculate the indefinite integral, where ${a}_{0},{a}_{1},{a}_{2},{a}_{3},{a}_{4}$ are constants:

$\int \frac{{a}_{0}{x}^{4}+{a}_{1}{x}^{3}+{a}_{2}{x}^{2}+{a}_{3}x+{a}_{4}}{{x}^{2}+1}\text{d}x$

Exercise 2. Given that (x+1) is a factor of the polynomial (x3+1). Calculate the indefinite integral:

$\int \frac{3{x}^{2}-2x+1}{{x}^{3}+1}\text{d}x$

Exercise 3. Use the substitution $t=\mathrm{tan}\frac{\theta }{2}$ in order to calculate the indefinite integral:

$\int \frac{\text{d}\theta }{\mathrm{sin}\theta -\mathrm{cos}\theta }$

Exercise 4. Use a table of integrals in order to calculate the indefinite integral:

$\int \frac{\text{d}x}{{\left(3x-1\right)}^{2}\sqrt{x+2}}$

Previous topic: page 2 Methods

Next topic: page 4 Simple Differential Equations