How do you integrate #int x5^x dx# using integration by parts?

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Answer 1 · 2016-04-19T22:37:11

#intx(5^x)dx=(x(5^x))/ln5-5^x/(ln5)^2#

#intudv=uv-intvdu#

In the case of

#intx(5^x)dx#

We set

#u=x" "=>" "(du)/dx=1" "=>" "du=dx#

#dv=5^xdx" "=>" "intdv=int5^xdx" "=>" "v=5^x/ln5#

We plug these values in to see that

#intx(5^x)dx=x(5^x/ln5)-int5^x/ln5dx#

#=(x(5^x))/ln5-1/ln5int5^xdx#

#=(x(5^x))/ln5-1/ln5(5^x/ln5)+C#

#=(x(5^x))/ln5-5^x/(ln5)^2#