Question

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

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ I=\int \frac{dx}{x\left\{6{\left(\log x\right)}^2+7\log x+2\right\}} \\ \mathrm{Putting}\log x=t \\ \Rightarrow \frac{1}{x}dx=dt \\ \therefore I=\int \frac{dt}{6t^2+7t+2} \\ =\int \frac{dt}{\left(3t+2\right)\left(2t+1\right)} \\ \mathrm{Let}\frac{1}{\left(3t+2\right)\left(2t+1\right)}=\frac{A}{3t+2}+\frac{B}{2t+1} \\ \Rightarrow \frac{1}{\left(3t+2\right)\left(2t+1\right)}=\frac{A\left(2t+1\right)+B\left(3t+2\right)}{\left(3t+2\right)\left(2t+1\right)} \\ \Rightarrow 1=A\left(2t+1\right)+B\left(3t+2\right) \\ \mathrm{Putting}2t+1=0 \\ \Rightarrow t=-\frac{1}{2} \\ 1=0+B\left(3\times -\frac{1}{2}+2\right) \\ \Rightarrow 1=B\left(\frac{1}{2}\right) \\ \Rightarrow B=2 \\ \mathrm{Putting}3t+2=0 \\ \Rightarrow t=-\frac{2}{3} \\ 1=A\left(2\times -\frac{2}{3}+1\right)+0 \\ \Rightarrow 1=A\left(-\frac{4}{3}+1\right) \\ \Rightarrow 1=A\left(-\frac{1}{3}\right) \\ \Rightarrow A=-3 \\ \therefore I=\int \left(-\frac{3}{3t+2}+\frac{2}{2t+1}\right)dt \\ =-3\frac{\log \left|3t+2\right|}{3}+2\frac{\log \left|2t+1\right|}{2}+C \\ =-\log \left|3t+2\right|+\log \left|2t+1\right|+C \\ =\log \left|\frac{2t+1}{3t+2}\right|+C \\ =\log \left|\frac{2\log x+1}{3\log x+2}\right|+C \end{gathered}$$