How do you find the integral of #(u-4)/ [u^2 +1]#? Calculus Techniques of Integration Integration by Substitution 1 Answer Lucy Apr 3, 2018 #1/2In abs(u^2+1)-4tan^(-1)(u^2+1)+c# Explanation: #int(u-4)/(u^2+1) du# =#intu/(u^2+1)-(4)/(u^2+1) du# =#1/2In abs(u^2+1)-4tan^(-1)(u^2+1)+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 4800 views around the world You can reuse this answer Creative Commons License