Question

1/x(4+x^6)的原函数大学数学不定积分.

Answer 1

∫1/x(4+x^6)dx

=∫x^2dx/[x^3(4+x^6)]

=1/3∫d(x^3)/[x^3(4+x^6)],令u=x^3

=1/3∫du/[u(4+u^2)]

=1/12∫du[1/u-u/(4+u^2)]

=1/12[∫du/u-0.5∫d(u^2)/(4+u^2)]

=1/12[ln|u|-0.5ln(4+u^2)]+C

=1/24[2ln|x^3|-ln(4+x^6)]+C