How do you integrate #y = (5x + 3)^6#?

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Answer 1 · 2016-11-01T01:55:53

#=1/35*(5x+3)^7+C#

So you've got...

#int(5x+3)^6dx#

Now, make:

#u=5x+3#

Which means that:

#(du)/(dx)=5#

And as a result:

#(dx)/(du)=1/5#

#dx=1/5du#

Now, transform:

#int(5x+3)^6dx# into #1/5intu^6du#

Integrate:

#1/5intu^6du#

#=1/5*u^7/7+C#

#=1/35*u^7+C#

#=1/35*(5x+3)^7+C#

#=int(5x+3)^6dx#