What is the integral of the expression #dy = (7x-2)^5 dx#? Help!?

1 Answer
Aug 23, 2017

#y = 1/42(7x-2)^6+C#

Explanation:

Given: #dy = (7x-2)^5 dx#

Then

#y = int (7x-2)^5dx#

Let #u = 7x-2#, then #du = 7dx# rewriting to that it can be easily substituted #dx = 1/7du#

#y = 1/7int (u)^5du#

#y = 1/42u^6+C#

Reverse the substitution:

#y = 1/42(7x-2)^6+C#