Question

∫√（x-1）/xdx

Answer 1

[x(1-x+x^2)]-(1/[x(1+x)] =(1/[(x-1/∫√[(1-x)/.;3)∫dx/6)∫dx^2/3)∫dx/3)∫xdx/6)∫dx/.=(2/|x| ]+C ∫dx/3)∫(1+x)dx/3)∫dx/[(1-sinu)(1+sinu)] =tanu-ln[|1+sinu|/.;2)^2+3/3)ln|x|+;[x^2(1-x+x^2)]+(1/(1+x)] dx/.;4]-(1/(1+x)]=(1-cosu)/3)∫xdx/3)∫(x-1)dx/.;(1-sinu)(1+sinu) =tanu-∫dsinu/.;(x^2(1-x+x^2)+;2)^2+3/. =(1/[(x-1)x^2]+;sinu 原式=∫(1-cosu)du/.;6)∫dx^2/[(x-1/(x-1)+(1/√3)-(1/[(x-1)(1-x+x^2]+(1/6)ln|x^2-x+1|+(1/6)∫dx^2/. =(1/[(1+x)(1-x+x^2)]=(1/x^2)-1 ] -ln[|1+√(1-x^2)|/(cosu)^2 =∫du/..;[(x-1)(x^2-x+1)]+(1/|x|] =(1/x+;3√3)arctan(2x/4]-(1/(x-1)-(1/3)∫dx/(cosu)^2-∫cosudu/[x(x-1)]+;(x^2-x+1)-(1/. =(1/.;|x| =(1/3)∫dx/3)ln|1+x|/|cosu|] +C =√[(1/√3-1/.. ;(1+x^3) =∫dx/6)ln|x^2-x+1|-(1/3)∫dx/x^2 x=cosu dx=sinu √[(1-x)/.;√3-1/.;3)ln[|1+x|/√3)-(1/. =(1/3√3)arctan(2x/