You can't do an integral involving "ln" unless you know what "ln" means! And surely when you learned about ln(x) you learned that \(\displaystyle ln(x^a)= a ln(x)\), one of the important properties of the logarithm. So \(\displaystyle \int ln(\sqrt[n]{x} dx= \int ln(x^{1/n})dx= \int \frac{1}{n} ln(x)dx= \frac{1}{n}\int ln(x)dx\) since 1/n is a constant with respect to x.
For the second one, exactly how have you defined "ln(x)"? Probably as the inverse function to \(\displaystyle f(x)= e^x\). That is, if \(\displaystyle y= ln(x)\) then \(\displaystyle x= e^y\). Then \(\displaystyle \frac{dx}{dy}= \frac{de^y}{dx}= e^y\). So \(\displaystyle \frac{d ln(x)}{dx}= \frac{dy}{dx}= \frac{1}{\frac{dx}{dy}}= \frac{1}{e^y}\) and, since \(\displaystyle e^y= x\), \(\displaystyle \frac{d ln(x)}{dx}= \frac{1}{x}\). Of course, since ln(x) is only defined for x> 0, we have to require that x> 0 here.
More generally, if u(x) is a function of x that may be either positive or negative, we have to take the absolute value of u so that ln(|u|) will have meaning. By the "chain rule", \(\displaystyle \frac{d ln(|u|)}{dx}= \frac{ln(|u|)}{du}\frac{d|u|}{dx}= \frac{1}{|u|}\frac{d|u|}{dx}\) which may be \(\displaystyle \frac{du}{dx}\) if u is positive or \(\displaystyle -\frac{du}{dx}\) if u is negative. But |u| is also u if u is positive or -u if u is negative so the signs cancel and we can simply write \(\displaystyle \frac{d ln(|u|)}{dx}= \frac{1}{u}\frac{du}{dx}\).
It is also perfectly valid to define \(\displaystyle ln(x)= \int_1^x \frac{1}{t} dt\) and then define \(\displaystyle exp(x)= e^x\) as the inverse function. In that case \(\displaystyle \frac{d ln(x)}{dx}= \frac{1}{x}\) immediately from the "Fundamental Theorem of Calculus".
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