What is #int g(x) = (5-6x^3) / x dx#?

1 Answer
Nov 11, 2015

#int(5-6x^3)/xdx = 5ln(x) - 2x^3 + c#

Explanation:

Umm, you kinda mixed up the notation there buddy.

#g(x) = int(5-6x^3)/xdx#

Means which function is #g(x)#, if it is that integral

Whereas

#int(g(x))dx = (5 - 6x^3)/xdx#

Seems to imply that the integral of #g(x)# is that, so which function is #g(x)#

I'm guessing you mean the first one, which the way you wrote isn't technically wrong but it's a bit confusing. In that case, we have

#int(5-6x^3)/xdx = int(5/x - 6x^2)dx#
#int(5/x - 6x^2)dx = int(5/x)dx - int(6x^2)dx#
#int(5/x)dx - int(6x^2)dx = 5intdx/x - 6intx^2dx#
#5intdx/x - 6intx^2dx = 5(ln(x)) - 6(x^3/3) + c#
#int(5-6x^3)/xdx = 5ln(x) - 2x^3 + c#