Question

помогите, пожалуйста

Answer 1

$intlimits^9_1 frac{x-1}{sqrt{x}}, dx = intlimits^9_1 (sqrt{x}-frac{1}{sqrt{x}})dx= intlimits^9_1 (x^{frac{1}{2}}-x^{-frac{1}{2}}) , dx =\\=(frac{2x^{frac{3}{2}}}{3} -frac{x^{frac{1}{2}}}{frac{1}{2}} )Big |_1^9=frac{2}{3}(sqrt{9^3}-1)-2(sqrt9-1)=\\=frac{2}{3}(3^3-1)-2(3-1)=frac{52}{3}-4=frac{40}{3}=13frac{1}{3}$

Answer 2

$\ int_{1}^{9}{x-1over sqrt{x} }{mathrm{d}x}=|u=x-1, du=dx; dv=x^{-{1over2}}, v=2sqrt{x}|=\\ =(v*u)|_{1}^{9}-int_{1}^{9}v*du=(2sqrt {x}*(x-1))|_{1}^{9}-2int_{1}^{9}sqrt {x}{mathrm{d}x}=\\ =2(3(9-1)-1(1-1))-{4over3}x^{{3over2}}|_{1}^{9}=48-{4over3}(9^{{3over2}}-1)=48-{4over3}(27-1)=\\=48-{4*26over3}=8(6-{13over3})=8({5over3})={40over3}$