Evaluate: `int((1+logx)^2)/x\ dx`

Explanation

Solution:

Let `t=1+\log x \Rightarrow d t=\frac{d x}{x}`

`I=\int \frac{(1+\log x)^{2}}{x} d x`

=`\int \{t}^{2} \{dt}`

=`\frac{t^{3}}{3}+c` where c is the constant of integration.

=`\frac{(1+\log x)^{3}}{3}+c` where `t=1+\log x`