What is the Integral of 1 over 2 to the Natural Log of X?

[tmt1]

I have this integral

$$\int_{}^{}\frac{1}{{2}^{lnx}} \,dx$$

I'm not sure the best way to do it.

I tried u-substitution:

$u = {2}^{lnx}$ and thus $u = {x}^{ln2}$, therefore $du = ln2({n}^{ln2 - 1}) dx$. However, not sure how to proceed from there.

 

[MarkFL]

I would let:

\(\displaystyle u=\ln(x)\implies dx=e^u\,du\)

And you now have:

\(\displaystyle I=\int\left(\frac{e}{2}\right)^u\,du\)

Now, use the fact that:

\(\displaystyle \frac{d}{dv}\left(\frac{a^v}{\ln(a)}\right)=a^v\)

to finish. :)

 

[tmt1]

I would let:

\(\displaystyle u=\ln(x)\implies dx=e^u\,du\)

And you now have:

\(\displaystyle I=\int\left(\frac{e}{2}\right)^u\,du\)

Now, use the fact that:

\(\displaystyle \frac{d}{dv}\left(\frac{a^v}{\ln(a)}\right)=a^v\)

to finish. :)

So it would be $$\frac{{(\frac{e}{2})}^{lnx}}{ln(\frac{e}{2})} +C$$?

 

[MarkFL]

Yes, looks correct to me. :D