Verified answer

Ответ:

[tex]\displaystyle \int\limits^{2}_{1} {\frac{x}{\sqrt{x-1} } } \, dx =2\frac{2}{3}[/tex],

Пошаговое объяснение:

[tex]\displaystyle \int\limits^{2}_{1} {\frac{x}{\sqrt{x-1} } } \, dx =[/tex]
Пусть х-1 = t, тогда dx=dt и x = t+1
[tex]\displaystyle =\int\limits^{2}_{1} {\frac{t+1}{\sqrt{t} } } \, dt=\int\limits^{2}_{1} {\frac{t}{\sqrt{t} } } \, dt+\int\limits^{2}_{1} {\frac{1}{\sqrt{t} } } \, dt=\int\limits^{2}_{1} {t^{1-0,5}} \, dt+\int\limits^{2}_{1} {t^{-0,5}} \, dt=\int\limits^{2}_{1} {t^{1/2}} \, dt+\int\limits^{2}_{1} {t^{-1/2}} \, dt=\frac{2t^{3/2}}{3}| ^{2}_{1} +2t^{1/2}|^{2}_{1} =\frac{2}{3}\sqrt{t^3}|^{2}_{1} +2\sqrt{t}|^{2}_{1} =[/tex]
Вернёмся к замене:
[tex]\displaystyle =\frac{2}{3}*\sqrt{(x-1)^3}|^{2}_{1} +2*\sqrt{x-1}|^{2}_{1} =\frac{2}{3}*(\sqrt{(2-1)^3}-\sqrt{(1-1)^3})+2*(\sqrt{2-1}-\sqrt{1-1} )=\frac{2}{3}*(1-0)+2*(1-0)=\frac{2}{3}+2 = 2\frac{2}{3}[/tex]