This problem becomes very simple once you realize that #v^(-1)# is equivalent to #1/v#. Therefore the problem is:
#int_2^4 1/vdv#
Now our task is to find a function whose derivative is #1/v#. That's the natural log function, of course! All that's left is to evaluate it from #2# to #4#:
#int_2^4 1/vdv=[lnv]_2^4#
#=ln4-ln2#
Using the property of logs that states #lna-lnb=ln(a/b)#, this becomes:
#ln(4/2)=ln2~~0.693...#
Note
Normally #int1/xdx=lnabs(x)#. The reason for this is that #1/x# is defined for all #x# except #x=0#, but #lnx# is defined for only #x>0#. This means that while #1/x# is defined for negative numbers, its antiderivative is not. The simple solution is to use the absolute value of #x# rather than just plain #x#, making the natural log function defined for all #x# except#x=0#, which is what we want. However, in my answer I didn't use the absolute value bars because we were only dealing with positive values (#2# to #4#) and it wasn't necessary.
