## Thomas' Calculus 13th Edition

$\displaystyle \frac{1}{6}$
Applying th.7, let $u=g(y)=1+\sqrt{y}$ (a continuous function). Then, $\quad \left[\begin{array}{ll} u=1+\sqrt{y}, & du=g'(y)dy\\ u=1+y^{1/2} & du=\frac{1}{2}y^{-1/2}dy\\ & du=\frac{dy}{2\sqrt{y}} \end{array}\right]$, and, the borders change to $g(1)=1+1=2$ and $\quad g(4)=1+2=3$ $\displaystyle \int_{1}^{4}\frac{dy}{2\sqrt{y}(1+\sqrt{y})^{2}}=\int_{2}^{3}\frac{1}{u^{2}}\cdot du =\int_{2}^{3}u^{-2}du$ $=\displaystyle \left[ \frac{u^{-2+1}}{-2+1} \right]_{2}^{3}$ $=\left[-u^{-1} \right]_{2}^{3}$ $=-\displaystyle \frac{1}{3}+\frac{1}{2}$ =$\displaystyle \frac{1}{6}$