5.5 The Substitution Rule/65

$\int _{1}^{2}x{\sqrt {x-1}}dx$

${\begin{aligned}u&=x-1\\u+1&=x\\[2ex]du&=1dx\\[2ex]du&=dx\end{aligned}}$

Failed to parse (unknown function "\du"): {\displaystyle \begin{align} \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 \\[2ex] &= \frac{2}{5} (u^\frac{5}{2} + \frac{2}{3} u^\frac{3}{2})\bigg| _{0}^{1} =\frac{2}{5} + \frac{2}{3} \\[2ex] &= \frac{16}{15}\\[2ex] \end{align} }