Integral of (x^5 + 1) / (x^6 + 6x) on x from 2 to 3.
\int_2^3 (x^5 + 1) / (x^6 +6x) dx
Not sure! The book only seems to give an equation for integrals in the form of du/u, which is integral 1/u*du = ln |u| + C. It also says that if u = f(x) THEN du = f'(x)?? Is that supposed to be a logical change? All of the integrals of this type are actually du/u in the book. This problem is not and is confusing the heck out of me.
The Attempt at a Solution
It appears that if I take u = x^6 + 6x and plug in the lower and upper limit, I will get a new limit that can be utilized in the correct answer. I already know what the answer is, but I want to figure out how to get to the answer:
1/6 ln (747/76)
The upper and lower limit when plugged into u are there: ln(747/76). I don't understand where the 1/6 is coming from and I don't understand how/why those two limits appears as a fraction.
Thanks for any help!