Answer
Convergent $\;,\;$ $\large\frac{32}{3}$
Work Step by Step
Let \[I=\int_{-2}^{14}\frac{dx}{\sqrt[4]{x+2}}\]
Since $-2$ is point of infinite discontinuity of integrand $\frac{1}{\sqrt[4]{x+2}}$
\[\Rightarrow I=\lim_{t\rightarrow -2^+}\int_{t}^{14}\frac{dx}{\sqrt[4]{x+2}}\;\;\;\ldots (1)\]
\[I=\lim_{t\rightarrow -2^+}\int_{t}^{14}(x+2)^{\frac{-1}{4}}dx\]
\[I=\lim_{t\rightarrow -2^+}\left[\frac{4}{3}(x+2)^{\frac{3}{4}}\right]_{t}^{14}\]
\[I=\lim_{t\rightarrow -2^+}\left[\frac{4}{3}(16)^{\frac{3}{4}}-\frac{4}{3}(t+2)^{\frac{3}{4}}\right]\]
\[I=\frac{4}{3}\times 8=\frac{32}{3}\]
Since limit on R.H.S. of (1) exists so $I$ is convergent and $I=\frac{32}{3}$.
