Integration of #(x-1)/(sqrtx(x-1))# dx? Calculus Introduction to Integration Integrals of Rational Functions 1 Answer Guillaume L. Aug 2, 2018 #I=2sqrtx+C#, #C in RR# Explanation: #I=int(x-1)/(sqrtx(x-1))dx# #=intcancel((x-1)/(x-1))^(=1)*1/sqrtxdx# #=2int1/(2sqrtx)dx# #=2sqrtx+C#, #C in RR# \0/ Here's our answer ! Answer link Related questions How do you integrate #(x+1)/(x^2+2x+1)#? How do you integrate #x/(1+x^4)#? How do you integrate #dx / (2sqrt(x) + 2x#? What is the integration of #1/x#? How do you integrate #(1+x)/(1-x)#? How do you integrate #(2x^3-3x^2+x+1)/(-2x+1)#? How do you find integral of #((secxtanx)/(secx-1))dx#? How do you integrate #(6x^5 -2x^4 + 3x^3 + x^2 - x-2)/x^3#? How do you integrate #((4x^2-1)^2)/x^3dx #? How do you integrate #(x+3) / sqrt(x) dx#? See all questions in Integrals of Rational Functions Impact of this question 2240 views around the world You can reuse this answer Creative Commons License