I am unsure on how to calculate the following 2 integrals

a) [1/(x^2)]*[e^(2/x)], with respect to xand

b) 1/(e^x + 1), with respect to x

Help with this will be greatly appreciated.

Here you have

a) \(\displaystyle \int{\left(\frac{1}{x^2}\right)e^{\frac{2}{x}}\,dx} = \int{e^{2x^{-1}}\,x^{-2}\,dx}\)

\(\displaystyle = -\frac{1}{2}\int{e^{2x^{-1}}(-2x^{-2})\,dx}\).

Now make the substitution \(\displaystyle u = 2x^{-1}\) so that \(\displaystyle du = -2x^{-2}\,dx\) and the integral becomes

\(\displaystyle -\frac{1}{2}\int{e^u\,du}\)

\(\displaystyle = -\frac{1}{2}e^u + C\)

\(\displaystyle = -\frac{1}{2}e^{2x^{-1}} + C\)

\(\displaystyle =-\frac{1}{2}e^{\frac{2}{x}} + C\).