Answer
\[\frac{1}{18}\ln(9x^2+6x+5)+\frac{1}{9}\tan^{-1}\left[\frac{1}{2}(3x+1)\right]+C\]
Work Step by Step
Let \[I=\int\frac{x+1}{9x^2+6x+5}dx\]
\[x+1=A\frac{d}{dx}[9x^2+6x+5]+B\]
\[x+1=A[18x+6]+B\]
Comparing like coefficients
\[18A=1\Rightarrow A=\frac{1}{18}\]
and
\[6A+B=1\Rightarrow B+\frac{1}{3}=1\Rightarrow B=\frac{2}{3}\]
\[\Rightarrow I=\frac{1}{18}\int\frac{18x+6}{9x^2+6x+5}dx+\frac{2}{3}\int\frac{dx}{9x^2+6x+5}\;\;\;\;\ldots (1)\]
Let \[I_1=\int\frac{18x+6}{9x^2+6x+5 }dx\]
Substitute $\;t=9x^2+6x+5\;\;\;\ldots (2)$
\[\Rightarrow dt=(18x+6)dx\]
\[I_1=\int\frac{dt}{t}=\ln t+C_1\]
Where $C_1$ is constant of integration
From (2)
\[I_1=\ln(9x^2+6x+5)+C_1\;\;\ldots (3)\]
Let \[I_2=\int\frac{dx}{9x^2+6x+5}\]
\[I_2=\frac{1}{9}\int\frac{dx}{x^2+\frac{2x}{3}+\frac{5}{9}}\]
\[I_2=\frac{1}{9}\int\frac{dx}{x^2+\frac{2x}{3}+\frac{5}{9} +\left(\frac{1}{3}\right)^2-\left(\frac{1}{3}\right)^2 }\]
\[I_2=\frac{1}{9}\int\frac{dx}{\left(x+\frac{1}{3}\right)^2+\left(\frac{2}{3}\right)^2}\]
\[I_2=\frac{1}{9}\left[\frac{1}{\left(\frac{2}{3}\right)}\tan^{-1}\left(\frac{x+\frac{1}{3}}{\frac{2}{3}}\right)\right]+C_2\]
Where $C_2$ is constant of integration
\[I_2=\frac{1}{6}\tan^{-1}\left(\frac{3x+1}{2}\right)+C_2\;\;\;\ldots (4)\]
Using (3) and (4) in (1)
\[I=\frac{1}{18}\ln(9x^2+6x+5)+\frac{1}{9}\tan^{-1}\left[\frac{1}{2}(3x+1)\right]+C\]
Where $C=C_1+C_2$
Hence
\[I=\frac{1}{18}\ln(9x^2+6x+5)+\frac{1}{9}\tan^{-1}\left[\frac{1}{2}(3x+1)\right]+C\]
