Question

Evaluate : $\int \frac{1+\log x}{x(2+\log x)(3+\log x)}dx$

Answer 1

Let $I=\int \frac{1+\log x}{x(2+\log x)(3+\log x)}dx$
Put $1+\log x=t\Rightarrow \frac{1}{x}dx=dt$
$I=\int \frac{t}{(1+t)(2+t)}dt$ .... (i)
Let $\frac{t}{(1+t)(2+t)}=\frac{A}{1+t}+\frac{B}{2+t}$
$t=A(2+t)+B(1+t)$ .... (ii)
Putting $t=-1$ in equation (ii), we get
$-1=A\Rightarrow A=-1$
Putting $t=-2$ in equation (ii), we get
$-2=-B\Rightarrow B=2$
Then equation (i), becomes
$I=\int [\frac{-1}{1+t}+\frac{2}{2+t}]dt$
$=-\int \frac{1}{1+t}dt+2\int \frac{1}{2+1}dt$
$=-\log |t+1|+2\log |t+2|+c$
$=2\log |\log x+3|-\log |\log x+2|+c$
$=\log |(\log x+3{)}^2|-\log |(\log x+2)|+c$
$\therefore \int \frac{1+\log x}{x(2+\log x)(3+\log x)}dx=\log \left|\frac{(\log x+3{)}^2}{(\log x+2)}\right|+c$