What is the integral of #int dx/(1+x^2) # from -1 to 1? Calculus Introduction to Integration Definite and indefinite integrals 1 Answer Maharshi Feb 26, 2016 #pi/2# Explanation: #int dx/(1+x^2), [-1,1] = tan^-1 (x), [-1,1]# #= tan^-1(1) - tan^-1 (-1)# #= pi/2# 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 1357 views around the world You can reuse this answer Creative Commons License