definite integration

(a) ∫ dx/(1 + x^2) [1/√3, √3]

= arctan x | [1/√3, √3]

= π/3 - π/6

= π/6

(b) Let u = 1/x, du = -dx/x^2

x = 1/√3, u = √3; x = √3, u = 1/√3

∫ dx/(1 + x^2)(1 + x^3) [1/√3, √3]

= - ∫ du/[(u^2)(1 + (1/u)^2)(1 + (1/u)^3)] [√3, 1/√3]

= ∫ u^3 du/[(1 + u^2)(1 + u^3)] [1/√3, √3]

= ∫ x^3 du/[(1 + x^2)(1 + x^3)] [1/√3, √3]

So, 2 ∫ dx/(1 + x^2)(1 + x^3) [1/√3, √3] = ∫ du/[(1 + x^2)] [1/√3, √3]

∫ dx/(1 + x^2)(1 + x^3) [1/√3, √3] = π/12