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

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

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

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

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

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