Answer
$\displaystyle \frac{1}{18}\ln\left(9x^{2}+6x+5\right)+\frac{1}{9}\tan^{-1}\left[\frac{1}{2}(3x+1)\right]+C$
Work Step by Step
$9x^{2}+6x+5=0$
$b^{2}-4ac=36-180$, prime... no partial fractins here.
$9x^{2}+6x+5=9x^{2}+6x+1 +4=(3x+1)^{2}+2^{2}$
$I=\displaystyle \int\frac{x+1}{9x^{2}+6x+5}dx=$
$\left[\begin{array}{llll}
u=3x+1 & ... & x=\frac{u-1}{3} & x+1=\frac{u+2}{3}\\
du=3dx & & dx=\frac{du}{3} &
\end{array}\right]$
$=\displaystyle \frac{1}{3}\cdot\frac{1}{3}\int\frac{u+2}{u^{2}+2^{2}}du$
$=\displaystyle \frac{1}{9}\int\frac{u}{u^{2}+4}du+\frac{1}{9}\int\frac{2}{u^{2}+2^{2}}du$
1st integral: substitute $\displaystyle \left[\begin{array}{l}
t=u^{2}+4\\
dt=2udu
\end{array}\right], \int\frac{u}{u^{2}+4}du=\frac{1}{2}\int\frac{dt}{t}$
2nd integral: Table of integrals, 17. $\displaystyle \int\frac{du}{a^{2}+u^{2}}=\frac{1}{a}\tan^{-1}\frac{u}{a}+C$
$I=\displaystyle \frac{1}{9}\cdot\frac{1}{2}\ln\left(u^{2}+4\right)+\frac{2}{9}\cdot\frac{1}{2}\tan^{-1}\left(\frac{1}{2}u\right)+C$
$=\displaystyle \frac{1}{18}\ln\left(9x^{2}+6x+5\right)+\frac{1}{9}\tan^{-1}\left[\frac{1}{2}(3x+1)\right]+C$
