5.5 The Substitution Rule/65

$\int _{1}^{2}x{\sqrt {x-1}}dx=\int _{0}^{1}u+1{\sqrt {u}}du=\int _{0}^{1}(u+1)({\sqrt {u}})=\int _{0}^{1}u^{\frac {3}{2}}+{\sqrt {u}}du={\frac {2}{5}}U^{\frac {5}{2}}+{\frac {2}{3}}U^{\frac {3}{2}}|_{0}^{1}={\frac {2}{5}}+{\frac {2}{3}}={\frac {16}{15}}$

Failed to parse (unknown function "\begin{align}"): {\displaystyle \begin{align} u &= x-1 ———> u+1 &= x \\[2ex] du &= 2 dx \\[2ex] \frac{1}{2} du &= dx \end{align} }