Question

x(x^2+1)分之一的不定积分

Answer 1

∫{1/[x(x^2+1)]}dx=∫{x/[x^2(1+x^2)]}dx=(1/2)∫{1/[x^2(1+x^2)]}d(x^2)
=(1/2)∫[1/x^2-1/(1+x^2)]dx^2
=(1/2)[lnx^2-ln(1+x^2)]+C
=ln√[x^2/(1+x^2)]+C