\(\int \frac{1}{x\left[6(\log x)^2+7 \log x+2\right]} d x\) is

\(I=\int \frac{1}{x\left[6(\log x)^2+7 \log x+2\right]} d x\)
\(=\int \frac{d t}{6 t^2+7 t+2} \quad\left[\begin{array}{l}\because \text { put } t=\log x \\ d t=\frac{1}{x} d x\end{array}\right]\)
\(=\int \frac{d t}{(3 t+2)(2 t+1)}\)
Now, \(\frac{1}{(3 t+2)(2 t+1)}=\frac{A}{(3 t+2)}+\frac{B}{(2 t+1)}\)
\(=\frac{A(2 t+1)+B(3 t+2)}{(3 t+2)(2 t+1)}\)
\(\Rightarrow \quad 1=A(2 t+1)+B(3 t+2)\)
Now, on putting \(t=\frac{-1}{2}\)
\(\Rightarrow \quad 1=0+B\left[3\left(-\frac{1}{2}\right)+2\right]\)
\(\Rightarrow \quad B=2\)
and put \(t=\frac{-2}{3}\)
\(\Rightarrow \quad 1=A\left(2 \times\left(-\frac{2}{3}\right)+1\right)+0 \Rightarrow 1=A\left(-\frac{4}{3}+1\right)\)
\(\Rightarrow \quad A=-3\)
\(\therefore \quad I=\int\left(\frac{-3}{3 t+2}+\frac{2}{2 t+1}\right) d t\)
\(=\frac{-3 \log |3 t+2|}{3}+2 \frac{\log |2 t+1|}{2}+C\)
\(=-\log |3 t+2|+\log |2 t+1|+C\)
\(=\log \left|\frac{2 t+1}{3 t+2}\right|+C\)
\(=\log \left|\frac{2 \log x+1}{3 \log x+2}\right|+C \quad[\because t=\log x]\)