1/x(1+x^2)的积分

法一：

令u=x^2,则du=2xdx

∫1/[x(x^2+1)]dx

=1/2·∫1/[u(u+1)]du

=1/2·∫[1/u-1/(u+1)]du

=1/2·∫1/u du-1/2·∫1/(u+1) d(u+1)

=1/2·lnu-1/2·ln(u+1)+C

=1/2·ln[u/(u+1)]+C

=1/2·ln[x^2/(x^2+1)]+C

法二：

∫1/[x(x^2+1)]dx

=∫x/[x^2(x^2+1)] dx

=1/2·∫1/[x^2(x^2+1)] d(x^2)

=1/2∫[1/x^2-1/(x^2+1)] d(x^2)

=1/2·[lnx^2-ln(x^2+1)]+C

=1/2·ln[x^2/(x^2+1)]+C

忘记了的骚操作