What is the Integral of tan^-1x/(1+x^2)? Calculus Introduction to Integration Integrals of Trigonometric Functions 1 Answer Eddie Aug 3, 2016 = 1/2 (tan^-1 x )^2 + C Explanation: int \ tan^-1x/(1+x^2) \ dx = int \ tan^-1 x * d/dx (tan^(-1) x) \ dx = int d/dx (1/2(\ tan^-1 x )^2 ) \ dx = 1/2 (tan^-1 x )^2 + C Answer link Related questions How do I evaluate the indefinite integral intsin^3(x)*cos^2(x)dx ? How do I evaluate the indefinite integral intsin^6(x)*cos^3(x)dx ? How do I evaluate the indefinite integral intcos^5(x)dx ? How do I evaluate the indefinite integral intsin^2(2t)dt ? How do I evaluate the indefinite integral int(1+cos(x))^2dx ? How do I evaluate the indefinite integral intsec^2(x)*tan(x)dx ? How do I evaluate the indefinite integral intcot^5(x)*sin^4(x)dx ? How do I evaluate the indefinite integral inttan^2(x)dx ? How do I evaluate the indefinite integral int(tan^2(x)+tan^4(x))^2dx ? How do I evaluate the indefinite integral intx*sin(x)*tan(x)dx ? See all questions in Integrals of Trigonometric Functions Impact of this question 44974 views around the world You can reuse this answer Creative Commons License