integrating fraction/completing the square Does anyone know how to integrate the following?
$\frac{dx}{9x^2 + 6x + 17}$
I have been trying for ages and cannot get an answer anywhere close to the answer I get on maple?

Without many words and since you've already been given some ideas. We're looking for an arctangent (general set up for this particular case of integrals), so try to justify each step:
$$9x^2+6x+17=9\left(x^2+\frac{2}{3}x\right)+17=9\left(x+\frac{1}{3}\right)^2+16=$$
$$16\left[1+\left(\frac{3}{4}x+\frac{1}{4}\right)^2\right]\;\;,\;\;\text{and}\;\;\left(\frac{3}{4}x+\frac{1}{4}\right)'=\frac{3}{4}\implies$$
$$\int\frac{dx}{9x^2+6x+17}=\frac{1}{16}\int\frac{dx}{1+\left(\frac{3}{4}x+\frac{1}{4}\right)^2}=\frac{1}{12}\int\frac{\frac{3}{4}dx}{1+\left(\frac{3}{4}x+\frac{1}{4}\right)^2}=$$
$$=\frac{1}{12}\arctan\left(\frac{3}{4}x+\frac{1}{4}\right)+K$$
