Question

Value of ${\int }_1^5(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{(x-1)}})dx$ is

Answer 1

$\begin{array}{c}{\int }_1^5\left(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}\right)dx\\ x=1=t^2\\ {\int }_0^5\left(\left(t+1\right)+\left|t-1\right|\right)2tdt\frac{dx}{dt}=2t\\ {\int }_0^{1}2t\left[\left(t+1\right)+\left(1-t\right)\right]dt+{\int }_1^{0}2t\left[\left(t+1\right)+\left(t-1\right)\right]dt\\ =\frac{4}{2}{\left[t^2\right]}_0^1+{\int }_1^2{\left(2t\right)}^{2}dt\\ =2+4{\int }_1^2{t}^{2}dt\\ =2+\frac{4}{3}\left[7\right]\\ =\frac{2+28}{3}=\frac{34}{3}\end{array}$