Answer
$I=\frac{4097}{45}$
Work Step by Step
$
I=\displaystyle\int_0^1{\left( 1+\sqrt{x} \right) ^8dx}\\$
$\begin{matrix}
z=1+\sqrt{x}& z-1=\sqrt{x}\\
\frac{dz}{dx}=\frac{1}{2\sqrt{x}}& 2\sqrt{x}dz=dx\\
\end{matrix}$
$I=\displaystyle\int_1^2{z^82\sqrt{x}\ dz}\\
I=2\displaystyle\int_1^2{z^8\left( z-1 \right) dz}\\
I=2\displaystyle\int_1^2{z^9-z^8\ dz}\\
I=2\left[ \frac{1}{10}z^{10}-\frac{1}{9}z^9 \right] _{1}^{2}\\
I=2\left( \left[ \frac{2^{10}}{10}-\frac{2^9}{9} \right] -\left[ \frac{1}{10}-\frac{1}{9} \right] \right) \\
I=\frac{4097}{45}
$
