I = ∫ $\frac{\mathit{l}\mathit{o}\mathit{g}\mathit{x}}{\mathit{x}}$ dx
Substitute log x = t
Differentiating w.r.t. x, we get
$\frac{1}{\mathit{x}}$ dx = dt
On substitution, we get
I = ∫ $\frac{\mathit{l}\mathit{o}\mathit{g}\mathit{x}}{\mathit{x}}$ dx = ∫ t . dt = $\frac{{\mathit{t}}^2}{2}$ + C = $\frac{{\left(\mathit{l}\mathit{o}\mathit{g}\mathit{x}\right)}^2}{2}$ + C