I'm trying to show that $\displaystyle \int_{-2}^{14}\frac{dx}{\sqrt[4]{x+2}}=

\frac{32}{3}$

Since the integrand $\displaystyle \frac{1}{\sqrt[4]{x+2}}$ is continuous on $\displaystyle (-2,14]$ I can define

$\displaystyle \int_{-2}^{14}\frac{dx}{\sqrt[4]{x+2}}=\lim_{t->-2^+}\int_{t}^{14}\frac{dx}{\sqrt[4]{x+2}}$

$\displaystyle =\lim_{t->-2^+}\int_{t}^{14}(x+2)^{-\frac{1}{4}}dx$

$\displaystyle =\lim_{t->-2^+}-\frac{4}{3\sqrt[4]{(x+2)^4}}\displaystyle{]_{x=t}^{x=14}}$

$\displaystyle =-\frac{1}{4}+\lim_{t->-2^+}\frac{4}{3\sqrt[4]{(t+2)^3}}$

I'm stuck on the limit. If I didn't know any better, I would say that the integral diverges. I tried writing $\displaystyle \lim_{t->-2^+}\frac{4\sqrt[4]{(t+2)^3}}{3(t+2)^3}$ so I could use L-hopitals rule, but it doesn't seem to go anywhere since the derivative of the numerator always give me a function of the form $\displaystyle \frac{c}{f(t)}$ where $\displaystyle f(-2)=0$. So what's the trick here?