Answer
$$\int_{0}^{1}(1+\sqrt{x})^{8}dx=\frac{4097}{45}$$
Work Step by Step
$$let\ t=1+\sqrt{x},\ x=(t-1)^{2},\ dx=2(t-1)dt$$
$$\int_{0}^{1}(1+\sqrt{x})^{8}dx=2\int_{1}^{2}t^{8}(t-1)dt$$
$$=2\int_{1}^{2}(t^{9}-t^{8})dt=2\left [\frac{t^{10}}{10}-\frac{t^{9}}{9}\right ]_{1}^{2}$$
$$=2\left [\frac{1024}{10}-\frac{512}{9}\right ]-2\left [\frac{1}{10}-\frac{1}{9}\right ]$$
$$=\frac{4097}{45}$$
