How do you integrate #int4x(x^2+3)^(-3) dx#? Calculus Introduction to Integration Definite and indefinite integrals 1 Answer Sasha P. Oct 18, 2015 #-1/(x^2+3)^2 + C# Explanation: #int4x(x^2+3)^(-3) dx = 2 int (x^2+3)^(-3) 2xdx =# #= 2 int (x^2+3)^(-3) (x^2+3)'dx = 2 int (x^2+3)^(-3) d(x^2+3) =# #= 2 (x^2+3)^-2/-2 +C = -1/(x^2+3)^2 + C# Answer link Related questions What is the difference between definite and indefinite integrals? What is the integral of #ln(7x)#? Is f(x)=x^3 the only possible antiderivative of f(x)=3x^2? If not, why not? How do you find the integral of #x^2-6x+5# from the interval [0,3]? What is a double integral? What is an iterated integral? How do you evaluate the integral #1/(sqrt(49-x^2))# from 0 to #7sqrt(3/2)#? How do you integrate #f(x)=intsin(e^t)dt# between 4 to #x^2#? How do you determine the indefinite integrals? How do you integrate #x^2sqrt(x^(4)+5)#? See all questions in Definite and indefinite integrals Impact of this question 1721 views around the world You can reuse this answer Creative Commons License