How do you find the indefinite integral of #int 1/sqrt(x+1)#? Calculus Introduction to Integration Integrals of Rational Functions 1 Answer Noah G Jan 12, 2017 Let #u = x + 1#. Then #du = dx#. #=int 1/sqrt(u)du# #= int u^(-1/2) du# #=2u^(1/2) + C# #=2sqrt(x + 1) + C# Hopefully this helps! 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 11737 views around the world You can reuse this answer Creative Commons License