Integration by Partial Fractions

> with(student):

Rational Functions: Case 0.

Recall: where p and q are polynomials is a rational function. R is a proper rational function if the degree of p is less than the degree of q. If the degree of p is greater than or equal to the degree of q then R is improper . Recall from high school that if R is improper then you can use long division to write R as the sum of a polynomial and a proper rational function.

Use long division to covert the following improper rational functions to a "proper" form:

> R1:=(x^3+2)/(x^2+x-1);convert(R1,parfrac,x);

> R2:=(x+1)/(x-1);convert(R2,parfrac,x);

> R3:=(2*x^4+x-1)/(x^2+1);convert(R3,parfrac,x);

Note: the integration techniques we will develop will apply only to proper rational functions so it is mandatory that you learn how to divide polynomials!

Find the following integrals by first converting the improper rational functions to a "proper" form:

> I1:=Int((x-1)/(x+1),x);value(I1);convert(integrand(I1),parfrac,x);

> I2:=Int((t^2+2*t+2)/(t+1),t);value(I2);convert(integrand(I2),parfrac,t);

> I3:=Int((2*x^4+x-1)/(x^2+1),x);value(I3);convert(integrand(I3),parfrac,x);

Now, suppose you were trying to integrate a proper rational function that looked like . This would be difficult unless you happened to notice that . Then it would be easy. The right hand side of the preceeding equation is called the partial fraction decomposition of and is the name of the game for what follows.

Case 1: The denominator of is a product of distinct linear factors .

Note: When I write in the following sections I am assuming that is a proper rational function . In cases when it is not, it is assumed that we would first use long division to write in its "proper" form and then apply the following techniques.

Consider the example in which the denominator factors into distinct linear factors as follows: . The partial fraction decomposition of R is obtained by writing

= and solving this equation for the constants A, B and C (see your text or the blackboard for details). Once decomposed, it is a trivial matter to find an antiderivative for R

> R:=(x^2+2*x+3)/(x^3-2*x^2-3*x);convert(R,parfrac,x);int(R,x);

e.g.

> I1:=Int(3/(x^2-4),x);value(I1);

note: The above answer is more politely written as .

e.g.

> I2:=Int(1/(x^2-a^2),x);value(I2);

note: the above antiderivative would normally be reported as .

Case 2: is a product of linear factors , some of which are repeated .

In case 2 we consider where q factors into linear factors, but some are repeated. Suppose, for instance, q contains 3 factors of the form . In the decomposition then we would need three terms of the form: where each is a constant. For example, to decompose you would solve the equation = for A, B, C, D.

e.g. find

> p:=9*x^2-20*x-2;q:=6*x^3-18*x^2+24;

> factor(q);

> convert(p/q,parfrac,x);

> int(p/q,x);

Case 3: contains distinct irreducible quadratic factors .

Three Preliminary e.g.'s

Big e.g.: Find (use substitution ).

> Int(1/(x^2+a^2),x)=int(1/(x^2+a^2),x)+C;

e.g. Find

> Int((x+2)/(x^2+4),x)=int((x+2)/(x^2+4),x);

e.g. Find

> R:=1/(x^2+x+1):I1:=Int(R,x);completesquare(R,x);value(I1);

The decomposition.

It can happen that a polynomial cannot be factored completely into linear factors. It may be that q contains quadratic factors that cannot be factored i.e. irreducible quadratic factors. Recall how quadratic expressions may be factored using the quadratic formula (below):

> p:=a*x^2+b*x+c;solutions:=solve(p=0,x);

> a*(x-solutions[1])*(x-solutions[2]);

> expand(%);

It follows from the above lines that a quadratic expression is reducible (over the Real numbers) if and only if > 0. To decompose when q contains distinct irreducible quadratic factors you must include a term of the form for each irreducible term . For example to decompose you would solve the equation = for A, B, C, D.

e.g.

> Int((x^2+2*x-1)/(x^4+x^3+x^2+x),x);

> factor(x^4+x^3+x^2+x);

> convert((x^2+2*x-1)/(x*(x+1)*(x^2+1)),parfrac,x);

> Int((x^2+2*x-1)/(x*(x+1)*(x^2+1)),x)=int((x^2+2*x-1)/(x*(x+1)*(x^2+1)),x);

Case 4: contains repeated irreducible quadratic factors.

This case is handled analogously to the case of repeated linear factors. For example, to decompose you would solve the equation for A, B, C, D, E

> R:=(1)/(x*(x^2+2)^2);convert(R,parfrac,x);Int(R,x)=int(R,x);

Some "rationalizing" substitutions

e.g. Find I1 = using substitution :

> I1:=Int(1/(x*sqrt(x+1)),x);

First try: let u = x + 1

> I2:=changevar(u=x+1,I1,u);

Second try: let w = or, equivalently, :

> I3:=changevar(w^2=x+1,I1,w);

> value(I3);ans:=value(I1);

> convert(ans,ln);

And the polite answer is:

e.g. Find I1 = using substitution :

> I1:=Int(x^3/((x^2+1)^(1/3)),x);

First try: u = :

> changevar(u=x^2+1,I1,u);

Second try: w = or

> I2:=changevar(w^3=x^2+1,I1,w);

> value(I2);value(I1);

e.g. Find I1 =

> I1:=Int(1/(sqrt(x)+x^(1/3)),x);

> I2:=changevar(u^6=x,I1,u);convert(integrand(I2),parfrac,u);

> value(I2);value(I1);