Question

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

Answer 1

We have,

$\begin{array}{c}I=\int \frac{1}{x\left(6I{n}^2\left(x\right)+7In\left(x\right)+2\right)}dx\\ LetsassumeIn\left(x\right)=y\\ \Rightarrow \frac{1}{x}dx=dy\\ NowsubstitutingtheabovevalueintoI,Weget\\ I=\int \frac{1}{6y^2+7y+2}dy\\ =\int \frac{1}{6y^2+4y+3y+2}dy\\ =\int \frac{1}{2y\left(3y+2\right)+3y+2}dy\\ =\int \frac{1}{\left(2y+1\right)\left(3y+2\right)}dy\\ Letsusepartialfractiondecompositiontechinque\\ Letsassume\\ \frac{1}{\left(2y+1\right)\left(3y+2\right)}=\frac{A}{2y+1}+\frac{B}{3y+2}\\ \Rightarrow A\left(3y+2\right)+B\left(2y+1\right)=1\\ \Rightarrow \frac{A}{2}=1\\ \therefore A=2\\ Similarlyaty=\frac{-2}{3},weget,\\ B=-3\\ So,I=\int \frac{2}{2y+1}dy-\int \frac{3}{3y+1}dy\\ =\int \frac{1}{2y+1}d\left(2y+1\right)-=\int \frac{1}{3y+2}d\left(3y+2\right)\\ =In\left(2y+1\right)-In\left(3y+2\right)\\ =In\left(\frac{2y+1}{3y+2}\right)+C\end{array}$

On putting the value of $y$, we get

$=\ln \left(\frac{2\log x+1}{3\log x+2}\right)+C$

Hence, this is the answer.