Intergrals calculus

Question

∫ from 1 to √3 6 / ((tan−1(𝑥))(1+𝑥^2) )𝑑𝑥

Answer 1

Let u = tan^-1(x). Then du = dx/(1 + x²). When x = 1, u = π/4 and when x = √3, u = π/3. So your integral becomes

∫[u = π/4 to π/3] 6/u du = 6 ln(|u|) [from u = π/4 to π/3] = 6 ln(π/3) - 6 ln(π/4) = 6 ln((π/3)/(π/4)) = 6 ln(4/3).