How to integrate int 2^lnx/xdx ?

3 Answers
Jun 23, 2018

2^log(x)/log(2)+C

Explanation:

Substituting log(x)=t then

1/xdx=dt
so we get

int2^tdt=2^t/log(2)+C

Aug 9, 2018

int frac{2^lnx}{x} dx= frac{2^lnx}{ln2} + C

Explanation:

int frac{2^lnx}{x} dx = int (2^(lnx))(1/x)dx

Use a u-substitution:

color(blue)(u = lnx)

color(blue)(du= 1/x dx)

int (2^(color(blue)(u)))du

= frac{2^(u)}{ln 2} + C

Substitute color(blue)(u=lnx) back in:

int frac{2^lnx}{x} dx= frac{2^lnx}{ln2} + C

Aug 9, 2018

2^lnx/ln2 +C

Explanation:

U-sub where u = ln(x)dx

u=lnxdx

du=1/xdx

int 2^udu
=
2^u/ln2 + C

sub ln(x) back in

2^lnx/ln2 + C