I need to evaluate the Inverse Trig Integral

Question

The integral is

$$\int \frac{x\,dx}{(3+2x+2x^2)}.$$

I'm stuck with breaking the denominator into $u^2+a^2$.

Answer 1

$$\int\frac{x \, dx}{2x^2 + 2x + 3} = \frac{1}{2}\int \frac{x \, dx}{x^2 + x + 3/2}$$

Now note that $x^2 + x + 3/2 = (x + 1/2)^2 + 5/4$.