设函数f(x)=(lnt)/(1+t^2)在1到x的定积分求fx-f(1/x)

Question

设函数f(x)=(lnt)/(1+t^2)在1到x的定积分求fx-f(1/x)

Answer 1

∫ f(x) dx = ln²x => f(x) = (2lnx)/x ∫ xf'(x² + 1) dx,令u = x² + 1,du = 2xdx => dx = du/(2x) = ∫ x * f'(u) * du/(2x) = (1/2)∫ f'(u) du = (1/2)f(u) + C = (1/2) * (2lnu)/u + C = [ln(x² + 1)]/(x² + 1) + C...