How do you integrate #(5x)/ (x-2)^(1/2) dx#? Calculus Techniques of Integration Integration by Substitution 1 Answer Tom Apr 11, 2015 #f(x)=5intx/(x-2)^(1/2)dx# Let's #u = x-2# #du = 1# #x = u +2# Now we have : #=5int(u+2)/u^(1/2) du# #=5intsqrt(u)du+10int1/sqrt(u)du# #=[10/3sqrt(u^3)]+20[sqrt(u)]+C# Substitute back #10/3(x-2)^(3/2)+20(x-2)^(1/2)+C# Answer link Related questions What is Integration by Substitution? How is integration by substitution related to the chain rule? How do you know When to use integration by substitution? How do you use Integration by Substitution to find #intx^2*sqrt(x^3+1)dx#? How do you use Integration by Substitution to find #intdx/(1-6x)^4dx#? How do you use Integration by Substitution to find #intcos^3(x)*sin(x)dx#? How do you use Integration by Substitution to find #intx*sin(x^2)dx#? How do you use Integration by Substitution to find #intdx/(5-3x)#? How do you use Integration by Substitution to find #intx/(x^2+1)dx#? How do you use Integration by Substitution to find #inte^x*cos(e^x)dx#? See all questions in Integration by Substitution Impact of this question 4475 views around the world You can reuse this answer Creative Commons License