1. ## Integral involving lnx

I have an integral i cant seem to figure out how to do.

$\int \frac {2^\ln x} { x } dx$

would i use u - substition like:

$u = {2^\ln x}$
Then,

$du = {e^\ln 2x$

Do, i then assign K = ln2

Then,

$\frac {e^k^x}{k}$

Then fill back in for k?

2. ## Re: Integral involving lnx

Originally Posted by icelated
I have an integral i cant seem to figure out how to do.

$\int \frac {2^\ln x} { x } dx$

would i use u - substition like:

$u = {2^\ln x}$
Then,

$du = {e^\ln 2x$

Do, i then assign K = ln2

Then,

$\frac {e^k^x}{k}$

Then fill back in for k?

$\int 2^{\ln x} \cdot \frac{1}{x} \, dx$

let $u = \ln{x}$ ...

3. ## Re: Integral involving lnx

wolfram gives me an answer 2^lnx / lnx how did they get there?

4. ## Re: Integral involving lnx

Originally Posted by icelated
wolfram gives me an answer 2^lnx / lnx how did they get there?
Are you sure you read that correctly?
Could it be $\frac{2^{\ln(x)}}{\ln(2)}~?$

5. ## Re: Integral involving lnx

yes , sorry thats what i meant. =(

6. ## Re: Integral involving lnx

Originally Posted by icelated
yes , sorry thats what i meant.
Find the derivative of that to see how it works.

7. ## Re: Integral involving lnx

the dirivitive is 1/x dx would i need to solve for x?

8. ## Re: Integral involving lnx

I think i get it. Du = 1/x dx

So, i would have 2^u du ?
then, integrate

9. ## Re: Integral involving lnx

Originally Posted by icelated
I think i get it. Du = 1/x dx

So, i would have 2^u du ?
then, integrate
that's it