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

1 Answer
Roy E.
Dec 26, 2016

Substitute #u=x+3# to get #intu^-2dx# and thence #-1/(x+3)+C#

Explanation:

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.

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