How do you integrate #int 1/(x^2+6x+9)# using substitution?

Question

Charles Anctil · Accepted Answer

Substitute #u=x+3# to get #intu^-2dx# and thence #-1/(x+3)+C# The denominator is a perfect square, namely #(x+3)^2# which greatly simplifies the process. Notice that the denominator is a parabola with its vertex at #(-3,0)#, so the #x#-axis is a tangent at that point. Consequently the graph of #1/(x^2+6x+9)# is just the graph of #1/x^2# transformed by a shift to the left by #3#. If the denominator had not been a perfect square, the substitution would have led either to something like #1/((x-a)(x-b))# or something like #1/((x-b)^2+a^2)#, taking you into the realm of logarithms and arctangents.

Christopher Agresta · Answer

undefined To integrate $ \int \frac{1}{x^2 + 6x + 9} $ using substitution, we can follow these steps:

1. Complete the square in the denominator: $ x^2 + 6x + 9 = (x + 3)^2 $.
2. Let $ u = x + 3 $, so $ du = dx $.
3. Rewrite the integral in terms of $ u $: $ \int \frac{1}{u^2} du $.
4. Integrate $ \frac{1}{u^2} $ with respect to $ u $ to get $ -\frac{1}{u} + C $.
5. Substitute back $ x + 3 $ for $ u $: $ -\frac{1}{x + 3} + C $.

Therefore, the integral of $ \frac{1}{x^2 + 6x + 9} $ using substitution is $ -\frac{1}{x + 3} + C $, where $ C $ is the constant of integration.