Here's the question you clicked on:
j814wong
Integral Question
\[\int\limits_{0}^{4} \frac{ dx }{ (x-2)^{2/3} }\]
You ARE going to have to cut it up at x = 2 and consider convergence on both sides.
Split it and take limit approaching 2 from left and right?
\[\int_0^4\frac{dx}{(x-2)^{\frac{2}{3}}}=\int_0^2\frac{dx}{(x-2)^{\frac{2}{3}}}+\int_2^4\frac{dx}{(x-2)^{\frac{2}{3}}}\] The integrand is undefined at x = 2, so you'll have to split up the integrals and take limits.
How do I take the integral of [1/(x-2)]^(2/3)
Simple exponent. \(\int [1/(x-2)]^{2/3}\;dx = \int [x-2]^{-2/3}\;dx = \dfrac{(x-2)^{1/3}}{1/3} + C\)
Gah... How did I miss that. Thanks.