How do you integrate #int 1/(3x)^2dx#?

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Answer 1 · 2017-02-07T21:49:33

#int1/(3x)^2dx=-1/(9x)+C#

Since #(3x)^2=3x*3x=9*x^2#

We can say

#int1/(3x)^2dx=int1/(9x^2)dx#

and since #intcf(x)dx=cintf(x)dx#

#=1/9int1/x^2dx#

and since #1/x^2=x^(-2)#

#=1/9intx^-2dx#

#intf'(x)dx=f(x)+C#

and

#(-x^(-1))'=-(-x^(-2))=x^(-2)#

#=-1/9x^-1+C=-1/(9x)+C#