Question

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

Answer 1

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.