How do you evaluate #\int 6x e ^ { 5x } d x #?

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Answer 1 · 2017-12-19T17:50:41

#(6xe^(5x))/5-6/25e^(5x)+c#

Take the integral

#int6xe^(5x)dx#

Factor out constants

#6intxe^(5x)dx#

For the integrand #xe^(5x)# integrate by parts,

#u=x#
#du=dx#

#dv=e^(5x)dx#
#v=e^(5x)/5#

#6intxe^(5x)dx=6((xe^(5x))/5-inte^(5x)/5dx)#

#inte^(5x)/5dx=e^(5x)/25#

#6intxe^(5x)dx=6((xe^(5x))/5-e^(5x)/25)+c#

#6intxe^(5x)dx=(6xe^(5x))/5-6/25e^(5x)+c#