$\int \frac{\log (x+1)-\log x}{x(x+1)} d x$ equals

$-\frac{1}{2}\left[\log \left(\frac{x+1}{x}\right)\right]^2+c$

$I=\int \frac{\log \left(\frac{x+1}{x}\right)}{\frac{(x+1)}{x}} . \frac{1}{x^2} d x$

Now put t = $\frac{x+1}{x}=1+\frac{1}{x}$    ∴ $dt=-\frac{1}{x^2} d x$

∴  $I=-\int \log t . \frac{1}{t} d t=-\frac{1}{2}(\log t)^2+c=-\frac{1}{2} \log \left[\left(\frac{x+1}{x}\right)^2\right]+c$

Hence (3) is the correct answer.