How do you integrate #int (x+5)^6# using substitution? Calculus Techniques of Integration Integration by Substitution 1 Answer Euan S. Jul 22, 2016 #int (x+5)^6dx = 1/7(x+5)^7 + C# Explanation: Let #u = x+5 implies du = dx# #int u^6du = u^7/7 + C# #therefore int (x+5)^6dx = 1/7(x+5)^7 + 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 2773 views around the world You can reuse this answer Creative Commons License