How do you find the definite integral of #(x^4 - 1)/( x^2 + 1) dx# from -5 to -2? Calculus Introduction to Integration Definite and indefinite integrals 1 Answer Alberto P. Nov 3, 2016 36 Explanation: #int_-5^-2(x^4-1)/(x^2+1)dx=int_-5^-2(x^2-1)dx=# #[x^3/3-x]_-5^-2=-8/3+2-(-125/3+5)=(-8+6+125-15)/3=108/3=36# 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 1536 views around the world You can reuse this answer Creative Commons License