How do you evaluate #intdx/(1+x^2)# from 0 to 1? Calculus Introduction to Integration Definite and indefinite integrals 1 Answer Jim H Apr 23, 2015 That is derivative of #tan^-1 x#, so #int_0^1 dx/(1+x^2) = tan^-1 x]_0^1 = tan^-1(1)-tan^-1(0)# #= pi/4 - 0 = pi/4# 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 1523 views around the world You can reuse this answer Creative Commons License