设f(x)的一个原函数是ln^2(x),则不定积分xf'(x^2+1)等于？求解题过程

Question

如题所述

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
做这种题最紧要是有技巧，直接求出f'(x² + 1)然后再积分未免有点笨。

Answer 2

∫f(x)dx=(lnx)^2
f(x)=[(lnx)^2]'=2lnx*(1/x)
f'(x)=2-2lnx*(1/x^2)
f'(x^2+1)=2-2ln(x^2+1)*(1/(x^2+1)^2)
∫xf'(x^2+1)dx=∫x*[2-2ln(x^2+1)*(1/(x^2+1)^2)dx
=x^2-∫2xln(x^2+1)dx/(x^2+1)^2
=x^2-∫ln(x^2+1)d(x^2+1)/(x^2+1)^2
=x^2+∫ln(x^2+1)d(1/(x^2+1))
=x^2+(1/(x^2+1))ln(x^2+1)-∫(x^2+1)dln(x^2+1)
=x^2+(1/(x^2+1))ln(x^2+1)-∫2xdx
=(1/(x^2+1))ln(1+x^2)+C