Answer
$\displaystyle \frac{1}{6}$
Work Step by Step
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 $
$=\displaystyle \left[ \frac{u^{-2+1}}{-2+1} \right]_{2}^{3}$
$=\left[-u^{-1} \right]_{2}^{3}$
$=-\displaystyle \frac{1}{3}+\frac{1}{2}$
=$\displaystyle \frac{1}{6}$
