How do you integrate #(5/sqrt x) dx#? Calculus Introduction to Integration Integrals of Rational Functions 1 Answer ali ergin Aug 4, 2016 #int 5/(sqrt(x))d x=10 sqrt x +C# Explanation: #int 5/(sqrt(x))d x=?# #"we can rearrange the term as;" # #int 5(x)^(-1/2)d x=5 int(x)^(-1/2) d x# #int int 5/(sqrt(x))d x=5(1/(-1/2+1)) x^(-1/2+1)# #int 5/(sqrt(x))d x=5*2*x^(1/2)# #int 5/(sqrt(x))d x=10 sqrt x +C# 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 13050 views around the world You can reuse this answer Creative Commons License