求∫1/[x（x^2+1）]dx的不定积分
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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
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
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