$\displaystyle \int h(x)f^1(x) * dx = \int$
and
$\displaystyle \int \frac{2}{3x^2 + 4x + 5} * dx = \int
$
The bottom does not factorise, so you can't use partial fractions.
That means you'll have to complete the square and use a trigonometric substitution.
$\displaystyle \int{\frac{2}{3x^2 + 4x + 5}\,dx} = \frac{2}{3}\int{\frac{1}{x^2 + \frac{4}{3}x + \frac{5}{3}}\,dx}$
$\displaystyle = \frac{2}{3}\int{\frac{1}{x^2 + \frac{4}{3}x + \left(\frac{2}{3}\right)^2 - \left(\frac{2}{3}\right)^2 + \frac{5}{3}}\,dx}$
$\displaystyle = \frac{2}{3}\int{\frac{1}{\left(x + \frac{2}{3}\right)^2 + \frac{11}{9}}\,dx}$
Now make the substitution $\displaystyle x + \frac{2}{3} = \sqrt{\frac{11}{9}}\tan{\theta} = \frac{\sqrt{11}}{3}\tan{\theta}$.
Note that $\displaystyle dx = \frac{\sqrt{11}}{3}\sec^2{\theta}\,d\theta$.
Also note that $\displaystyle \theta = \arctan{\left[\frac{3\sqrt{11}}{11}\left(x + \frac{2}{3}\right)\right]}$.
Then the integral becomes
$\displaystyle = \frac{2}{3}\int{\frac{1}{\left(\frac{\sqrt{11}}{3} \tan{\theta}\right)^2 + \frac{11}{9}}\,\frac{\sqrt{11}}{3}\sec^2{\theta}\, d\theta}$
$\displaystyle = \frac{2}{3}\int{\frac{\frac{\sqrt{11}}{3}\sec^2{\t heta}}{\frac{11}{9}\tan^2{\theta} + \frac{11}{9}}\,d\theta}$
$\displaystyle = \frac{2}{3}\int{\frac{\frac{\sqrt{11}}{3}\sec^2{\t heta}}{\frac{11}{9}(\tan^2{\theta} + 1)}\,d\theta}$
$\displaystyle = \frac{2\sqrt{11}}{11}\int{\frac{\sec^2{\theta}}{\s ec^2{\theta}}\,d\theta}$
$\displaystyle = \frac{2\sqrt{11}}{11}\int{1\,d\theta}$
$\displaystyle = \frac{2\sqrt{11}}{11}\theta + C$
$\displaystyle = \frac{2\sqrt{11}}{11}\arctan{\left[\frac{3\sqrt{11}}{11}\left(x + \frac{2}{3}\right)\right]} + C$.