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#