How to integrate #int 2^lnx/xdx# ?

Question

5584 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-06-23T08:58:16

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

Substituting #log(x)=t# then

#1/xdx=dt#
so we get

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

Answer 2 · 2018-08-09T22:58:43

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

#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#

Answer 3 · 2018-08-09T23:06:15

#2^lnx/ln2 +C#

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#