How do I evaluate #int5y ln (9y) dy#?

Question

1678 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-01-30T17:18:56

I would use integration by parts where you have:

#intf(y)*g(y)dy=F(y)*g(y)-intF(y)*g'(y)dy#

Where:

#F(y)=intf(y)dy#
#g'(y)# is the derivative of #g(y)#

In your case you can choose:

#f(y)=5y#
#g(y)=ln(9y)#

And so:

#int5y*ln(9y)dy=5y^2/2*ln(9y)-int5y^2/2*1/(9y)*9dy=#
#=5y^2/2*ln(9y)-int5y/2dy=#
#=5y^2/2*ln(9y)-5/4*y^2+c#