How do you evaluate the definite integral of 0 to 4 for #(5 / 3x +1) dx#? Calculus Introduction to Integration Integrals of Polynomial functions 1 Answer Konstantinos Michailidis Sep 14, 2015 It is #52/3# Explanation: It is #int_0^4 (5/3x+1)dx=int_0^4 (5/3x^2/2+x)'dx=[5x^2/6+x]_0^4=5/6*4^2+4=52/3# Answer link Related questions How do you evaluate the integral #intx^3+4x^2+5 dx#? How do you evaluate the integral #int(1+x)^2 dx#? How do you evaluate the integral #int8x+3 dx#? How do you evaluate the integral #intx^10-6x^5+2x^3 dx#? What is the integral of a constant? What is the antiderivative of the distance function? What is the integral of #|x|#? What is the integral of #3x#? What is the integral of #4x^3#? What is the integral of #sqrt(1-x^2)#? See all questions in Integrals of Polynomial functions Impact of this question 2033 views around the world You can reuse this answer Creative Commons License