How do you find the integral of int arctan(x) dx? Calculus Introduction to Integration Integrals of Trigonometric Functions 1 Answer Sasha P. Oct 20, 2015 I = xarctanx - 1/2 ln (x^2+1) + C Explanation: Integration By Parts: int udv = uv - int vdu u = arctanx => du = 1/(x^2+1) dx dv = dx => v=x I = int arctanx dx = xarctanx - int x* 1/(x^2+1)dx I = xarctanx - int (xdx)/(x^2+1) = xarctanx - int (1/2d(x^2+1))/(x^2+1) I = xarctanx - 1/2 int (d(x^2+1))/(x^2+1) I = xarctanx - 1/2 ln (x^2+1) + 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 1863 views around the world You can reuse this answer Creative Commons License