∫{(2x^2+1)/[x^2(x^2+1)]}dx的不定积分
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∫{(2x^2+1)/[x^2(x^2+1)]}dx
=∫2x^2/[x^2(x^2+1)]dx+∫1/[x^2(x^2+1)]dx
(前一项分子分母约去x^2,后一项利用1/[x^2(x^2+1)]=1/x^2-1/(x^2+1))
=2∫1/(x^2+1)dx+∫1/x^2 dx-∫1/(x^2+1)dx
=∫1/(x^2+1)dx+∫1/x^2dx
=arctanx - 1/x + C
C是任意常数.
=∫2x^2/[x^2(x^2+1)]dx+∫1/[x^2(x^2+1)]dx
(前一项分子分母约去x^2,后一项利用1/[x^2(x^2+1)]=1/x^2-1/(x^2+1))
=2∫1/(x^2+1)dx+∫1/x^2 dx-∫1/(x^2+1)dx
=∫1/(x^2+1)dx+∫1/x^2dx
=arctanx - 1/x + C
C是任意常数.
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