Evaluate: `int(sqrt(16+(logx)^2))/x\ dx`

Explanation

Let `{I}=int sqrt{frac{16+(log x)^{2}}{x}} {dx}`

Putting `log x=t`

`Rightarrow frac{1}{x} {dx}=d t `

`therefore I=int sqrt{16+t^{2}} d t `

`=int sqrt{4^{2}+t^{2}} d t`

`=frac{t}{2} sqrt{4^{2}+t^{2}}+frac{4^{2}}{2} log |t+sqrt{4^{2}+t^{2}}|+C `

`=frac{log x}{2} sqrt{16+(log x)^{2}}+8 log |log x+sqrt{16+(log x)^{2}}|+C`