Question

Evaluate $\int \frac{dx}{x\left\{6\right(\log x{)}^2+7\log x+2\}}$.

Answer 1

Let $\log x=t\Rightarrow \frac{1}{x}dx=dt$
$\therefore \int \frac{dx}{x\left\{6\right(\log x{)}^2+7\log x+2\}}$
$=\int \frac{dt}{6t^2+7t+2}$
$=\int \frac{dt}{(3t+2)(2t+1)}$

Let, $\frac{1}{(3t+2)(2t+1)}=\frac{A}{3t+2}+\frac{B}{2t+1}$
$1=A(2t+1)+B(3t+2)$

On putting t = -1/2 and t = -2/3 respectively, we get
$\therefore 1=A(2\times -2/3+1)$
$A=-3$
$1=B(3\times -\frac{1}{2}+2)$
$\Rightarrow B=2$
$\int \frac{dt}{(3t+2)(2t+1)}=\int \frac{-3}{3t+2}dt+\int \frac{2}{2t+1}dt$
$=-3.\frac{1}{3}\log |3t+2|+2.\frac{1}{2}\log |2t+1|+c$
$=-\log |3t+2|+\log |2t+1|+c$
$=\log \frac{|2t+1|}{|3t+2|}+c=\log \left|\frac{2\log x+1}{3\log x+2}\right|+c$