Indefinite Integration — Maths STD 12 Science — Question

Let $I=\int \frac{1+\log x}{x(2+\log x)(3+\log x)} d x$
Put
$\log x=t$
$\frac{1}{x} d x=d t$
$I=\int \frac{1+t}{(2+t)(3+t)} d t$
consider
$\frac{1+t}{(2+t)(3+t)}=\frac{A}{2+t}+\frac{B}{3+t}$
$(1+t)=A(3+t)+B(2+t)$
A=-1,B=2
$\begin{aligned} & \frac{1+t}{(2+t)(3+t)}=-\frac{1}{2+t}+\frac{2}{3+t} \\ & I=\int-\frac{1}{2+t} d t+\int \frac{2}{3+t} d t \\ & =-\log |(2+t)|+2 \log |(3+t)|+c \\ & =\log \left[\left|\frac{(3+t)^2}{2+t}\right|\right]+c \\ & =\log \left[\left|\frac{(3+\log x)^2}{2+\log x}\right|\right]+C\end{aligned}$