Integration by Completing the Square and Substitution

Recall that a rational function in mathematics is a function consisting of a polynomial divided by another polynomial, like:

The polynomial in the denominator should have a degree equal to at least one, meaning xx or a power thereof should feature somewhere below the line.

When it comes to the integration of a rational function, the difficulty of the problem depends on the actual expression you need to integrate. In the best-case scenario, it may boil down to a simple integration of polynomials. More complex examples will however lead to partial fractions. The first thing to check, however, is whether the degree of the numerator is greater than the degree of the denominator. If so, then you should use the polynomial long division calculator to simplify the problem.

Perform the following steps to integrate a rational function in the form L(x)/M(x)L(x)/M(x):

1. If the degree of the numerator L(x)L(x) is greater than the degree of the denominator M(x)M(x), perform long division on the polynomials to obtain WW and RR such that M(x)=L(x)W(x)+R(x)M(x)=L(x)W(x)+R(x).
2. If you end up with no remainder (i.e., R=0R=0), lucky you — there's only the polynomial WW to be integrated. Remember that the integral of xnxn is xn+1/(n+1)xn+1/(n+1), and you'll be done in no time.
3. If there is a remainder, you have to rewrite the rational function R(x)/M(x)R(x)/M(x) as the sum of partial fractions.
4. Once you have partial fractions, integrate them one by one. The formulas for integrating fractions of the form 1/(x+b)m1/(x+b)m and (px+q)/(x2+bx+c)n(px+q)/(x2+bx+c)n are to be found in any table of integrals.
5. Quadratic partial fractions of the form 1/(x2+bx+c)n1/(x2+bx+c)n require your special attention. We integrate them by using the completing the square method and then by substitution. We will explain this in more detail in the next section.

Here we discuss how to evaluate the integral of 1/(ax2+bx+c)n1/(ax2+bx+c)n using the method of completing the square and by substitution.

1. Rewrite x2+bx+cx2+bx+c as (x+b/2)2+(c−b2/4)(x+b/2)2+(c−b2/4), i.e., add and subtract the constant term b2/4b2/4.
2. Denoting c−b2/4=ac−b2/4=a and x+b/2=a​ux+b/2=a​u we get au2+aau2+a. Note that dx=a​dudx=a​du.
3. After this uu substitution, we arrive at the integral of 1/(1+u2)n1/(1+u2)n.
4. In the particular case of n=1n=1, the solution is equal to arctan(u)arctan(u), but for higher nn, the formula is much more complicated and can be found in integration tables.

As you can see, getting the correct answer to this calculus question requires quite a lot of computation. But don't get discouraged — as always, practice makes perfect. Going through a few exercises will certainly help, and soon you'll be the master of integration using completing the square method!