P5 edexcel 2005

x=\cosh u \\[br]dx=\sinh u du \\[br]\int_{1}^{2} xarcoshx dx = \int_{0}^{arcosh2} u\cosh u \sinh u du \\[br]=\frac{1}{2}\int_{0}^{arcosh2} usinh2u du \\[br]v=u,\frac{dv}{du}=1,\frac{dt}{du}=sinh2u,t=\frac{1}{2}cosh2u \\[br]\int_{1}^{2} xarcoshx dx =\frac{1}{2}[ [\frac{1}{2}ucosh2u]_{0}^{arcosh2} - \int_{0}^{arcosh2} \frac{1}{2} cosh2u du ] \\[br]=\frac{1}{4}(arcosh2)cosh(2arcosh2) - [\frac{1}{8}sinh2u]_{0}^{arcosh2}[br]=\frac{7}{4}ln(2+\sqrt{3})-[\frac{\sqrt{3}}{2}] \\[br]=\frac{7}{4}ln(2+\sqrt{3})-\frac{\sqrt{3}}{2} \\[br]