The question is $\displaystyle \int\frac{x+1}{9x^2+6x+5}$ I believe this is a partial fraction problem, but I don't know how to do it.
Any help is highly appreciated, thank you
Does the denominator factorise? If not, then partial fractions is not appropriate.
First look for simple substitutions. If you make the substitution $\displaystyle \displaystyle \begin{align*} u = 9x^2 + 6x + 5 \end{align*}$, then you require its derivative $\displaystyle \displaystyle \begin{align*} \frac{du}{dx} = 18x + 6 \end{align*}$ to be a factor. We can easily get that...
$\displaystyle \displaystyle \begin{align*} \int{\frac{x + 1}{9x^2 + 6x + 5}\,dx} &= \frac{1}{18}\int{ \frac{18x + 18}{9x^2 + 6x + 5}\,dx} \\ &= \frac{1}{18}\int{ \frac{18x + 6}{9x^2 + 6x + 5} + \frac{12}{9x^2 + 6x + 5} \, dx } \end{align*}$
So the first term can be easily integrated using that substitution. The second requires completing the square on the denominator, then it should either look familiar (as the derivative of an inverse trigonometric or hyperbolic function) or will require a trigonometric or hyperbolic substitution.
Playing with this a moment I see it's not partial fraction you want to use. Try completing the square in the denominator and making a substitution.
You want to target x^{2} + a^{2} in the denominator which is going to integrate to some sort of arctan for the constant over this term, and a ln for the x over this term.
Give it a shot.