how to integrate this

Question

$\int_{1}^{n}\frac{dt}{log (t)}$

Answer 1

Substitute log(t) = x

$\therefore$   $\frac{1}{t}dt$ = dx

$\therefore$  dt = e^x dx

t=1 $\Rightarrow$ x=0

t=n $\Rightarrow$ x=logn

So, integral 

= $\int_{0}^{logn}$  $\frac{e^{x}}{x}dx$

Now, if we expand the Mclaurin series for e^x, then denominator will cancel out. I think this will lead to infinity after integration.

See here