How do you integrate #int x*5^x# by integration by parts method?

1 Answer
Jun 11, 2017

#intx*5^x# #dx=#

Explanation:

#intx*5^x# #dx#

Let #f=x#, then #df=dx#

And #dg=5^x# #dx#, then #g=5^x /ln5#

#intx*5^x# #dx=(x*5^x)/ln5-1/ln5int5^x# #dx#

#=(x*5^x)/ln5-5^x/ln^2 5+"c"#

Taking out a factor of #1/ln^2 5# yields:

#intx*5^x# #dx=1/ln^2 5[(x*5^x)ln5-5^x]+"c"#

How to integrate #5^x# #dx#

#int5^x# #dx#

Let #u=5^x#

Then #du=5^xln5# #dx#

And #dx=1/(5^xln5)# #du#

#thereforeint5^x# #dx=int5^x xx 1/(5^xln5)# #du=1/ln5int1# #du=u/ln5+"c"#
#=5^x/ln5+"c"#