calculus u substitution

Question

∫from 1 to √3 6 / [(tan−1(𝑥))(1+𝑥2)] dx

Answer 1

Let u = tan^-1x Then du = [1 / (1+x²)] dx

When x = 1, u = tan^-1(1) = π/4 and when x = √3, u = tan^-1(√3) = π/3

So, the given integral is equivalent to ∫_{(from π/4 to π/3)} udu = (1/2)u²_{(from π/4 to π/3)}

(1/2) [π²/9 - π²/16]