Integration by Partial Fractions

#Calc-2#Math#Integrals We know the formula $\int \frac{1}{ax+b}dx=\frac{1}{a}\ln |ax+b|+C$ is true, so if we can convert an integral into the form of multiple of those types of fractions, we can easily integrate it.

Partial Fraction Expansion §

FOR PARTIAL FRACTION EXPANSION TO WORK THE NUMERATOR MUST HAVE A LOWER DEGREE THAN THE DENOMINATOR

If it does not, do long division so that it does.

Case 1: Denominator is a product of DISTINCT LINEAR factors §

• Distinct means they aren’t repeating
• Linear means they are of first degree

$\frac{x-8}{x^2-7x+10}=\frac{x-8}{(x-5)(x-2)}=\frac{A}{x-5}+\frac{B}{x-2}$ this is similar to how $\frac{1}{6}=\frac{A}{3}+\frac{B}{2}$ all we have to do is find A and B.

Do this by multiplying both sides of the equation by the denominator and solving the resulting system of equations for A and B

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

$\frac{10}{x^2(5-2x)}=\frac{10}{(x)(x)(5-2x)}$ In this case we do $\frac{10}{(x)(x)(5-2x)}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{5-2x}$ process for solving is the exact same

Case 3: Denominator contains distinct irreducible quadratics §

You should start to notice a pattern, which makes these pretty easy $\frac{2x^2-x+4}{x^3+4x}=\frac{2x^2-x+4}{x(x^2+4)}=\frac{A}{x}+\frac{Bx+C}{x^2+4}$ Solve the exact same ways

Case 4: Repeated Irreducible Quadratics §

You handle these just like handling repeated linear factors, except the numerator is itself a linear term.

$\frac{1}{x(x^2+1{)}^2}=\frac{A}{x}+\frac{Bx+C}{x^2+1}+\frac{Dx+E}{(x^2+1{)}^2}$ Solve same way

Example Combining all of the things §

$\frac{2x^3+5x-10}{x^2(x-3)(x^2+1{)}^3}=\frac{A}{x-3}+\frac{B}{x}+\frac{C}{x^2}+\frac{Dx+E}{x^2+1}+\frac{Fx+G}{(x^2+1{)}^2}+\frac{Hx+I}{(x^2+1{)}^3}$