OK, so the last post was a bit abstract, so it’s good to run through some examples here. Here are a few examples which illustrate some of the theory we discussed last time.

Example 1: A quadratic function in the denominator which can be written as the product of two linear terms

$\frac{3x-6}{x^2+5x+6}=\frac{3x-6}{(x+2)(x+3)}$

Now split this up into the sum of two terms:

$\frac{3x-6}{(x+2)(x+3)}=\frac{A}{x+2}+\frac{B}{x+3}$

multiply out the right hand side:

$\frac{3x-6}{(x+2)(x+3)}=\frac{A(x+3)+B(x+2)}{(x+2)(x+3)}=\frac{x(A+B)+3A+2B}{(x+2)(x+3)}$

and match the coefficient of $x$ and the constant part to find $A$ and $B$: $A+B=3$ and $3A+2B=-6$. The solution to this is $A=-12$ and $B=15$. Our final expression is:

$\frac{-12}{x+2}+\frac{15}{x+3}$

and if we need to integrate this we can use the normal techniques we have learnt up to now.

Example 2: An irreducible quadratic

The quadratic in the following can’t be factored into linear terms in $x$ (well, not using real numbers, anyway). We can complete the square in the quadratic term though:

$\frac{3x-6}{(x^2+6x+12)(x-3)(x+2)}=\frac{3x-6}{((x+3)^2+3)(x-3)(x+2)}$

we now write this as:

$\frac{3x-6}{((x+3)^2+3)(x-3)(x+2)}=\frac{A}{x-3}+\frac{B}{x+2}+\frac{Cx+D}{((x+3)^2+3)}$

Note that the quadratic term has not just a constant in the numerator, but a term $Cx+D$. This is very important to remember. To solve for $A,B, C$ and $D$, we have to recombine all terms:

$\frac{3x-6}{((x+3)^2+3)(x-3)(x+2)}=\frac{A}{x-3}+\frac{B}{x+2}+\frac{Cx+D}{((x+3)^2+3)}$

$=\frac{A(x+2)((x+3)^2+3)+B(x-3)((x+3)^2+3)+(Cx+D)(x-3)(x+2)}{((x+3)^2+3)(x-3)(x+2)}$

And match the coefficients of $x^2$, $x$ and the constant term between the left and the right expressions. Solve for these constants and then we can integrate each of the individual terms in the sum (if it is indeed an integral we want to be doing). The last term can be integrated by a $u$ substitution.

$\int \frac{Cx+D}{((x+3)^2+3)}dx$ with $u=x+3$:

$\int \frac{C(u-3)+D}{(u^2+3)}dx=\int \frac{C u}{(u^2+3)}+\frac{-3C+D}{(u^2+3)}du$

The first term can be integrated directly by noting that it is of the form $f(g(x))g'(x)$ and the second term can be solved by substituting $u=\sqrt{3}\tan\theta$.

Example 3 A term which appears multiple times (ie. to a power greater than 1) in the denominator

Here we have one of the factors to the power of 2:

$\frac{4x-3}{(x-3)(x+2)^2}$

Here $(x+2)$ appears twice (ie. it is to the second power). The partial fraction expansion of this is:

$\frac{4x-3}{(x-3)(x+2)^2}=\frac{A}{x-3}+\frac{B}{x+2}+\frac{C}{(x+2)^2}$

Again, solve for $A, B$ and $C$ by combining the fractions and matching coefficients.
Example 4: we have powers of an irreducible quadratic factor.

Again we must have multiple terms in the sum coming from the :

$\frac{3x-6}{((x+3)^2+3)^2(x-3)(x+2)}=\frac{A}{x-3}+\frac{B}{x+2}+\frac{Cx+D}{((x+3)^2+3)}+\frac{Ex+F}{((x+3)^2+3)^2}$

Again, combine the factors on the right into a single fraction and match coefficients. Solve for the coefficients of different powers of $x$, plug in the solution to the above expression and integrate.

 How clear is this post?