$\displaystyle \int_{-2}^{14} \dfrac{7}{(x + 2)^{1/4}} dx$

$\displaystyle \int_{b}^{14} \dfrac{7}{(x + 2)^{1/4}} dx$

$\displaystyle \int_{b}^{14} 7(x + 2)^{-1/4}$

$\displaystyle u = x + 2$

$\displaystyle du = dx$

$\displaystyle \int_{b}^{14} 7(u)^{-1/4}$

$\displaystyle = 7(\dfrac{u)^{3/4}}{3/4}$ evaluated at b and 14

$\displaystyle = 7((\dfrac{4}{3})u^{3/4}$ evaluated at b and 14

$\displaystyle = (\dfrac{28}{3})(u^{3/4}$ evaluated at b and 14

$\displaystyle = (\dfrac{28}{3})(((x + 2))^{3/4}$ evaluated at b and 14

$\displaystyle [(\dfrac{28}{3})(((14 + 2))^{3/4}] - [(\dfrac{28}{3})(((b + 2))^{3/4}]$

$\displaystyle \lim b \rightarrow \infty [(\dfrac{28}{3})(((14 + 2))^{3/4}] - [(\dfrac{28}{3})((((\infty) + 2))^{3/4}]$ ??

$\displaystyle \lim b \rightarrow \infty [(\dfrac{28}{3})(((16))^{3/4}] - [(\dfrac{28}{3})((((\infty) + 2))^{3/4}]$

$\displaystyle \lim b \rightarrow \infty [(\dfrac{28}{3})(8)] - [(\dfrac{28}{3})((((\infty) + 2))^{3/4}]$

$\displaystyle \lim b \rightarrow \infty [(\dfrac{224}{3})] - [\infty]$ ??