How do you integrate #f(x)=x^2/(3x-1)# using the quotient rule?

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Answer 1 · 2018-06-13T03:02:37

#(3x^2-2x)/(2x)^2#

If we have a function #f(x)# being defined as equal to #(h(x))/g(x)#, the Quotient Rule tells us that the derivative will be equal to

#(h'(x)g(x)-h(x)g'(x))/g(x)^2#

If we define our #f(x)# in such a way, then we know

#h(x)=x^2=>h'(x)=2x#

#g(x)=3x-1=>g'(x)=3#

We have everything we need to plug in, so let's do that now!

#(2x*(3x-1)-x^2(3))/(2x)^2#

Which simplifies to

#(6x^2-2x-3x^2)/(2x^2)#

Further simplifying to get

#f'(x)=bar( ul( |(3x^2-2x)/(2x)^2)|)#

Hope this helps!