Question

∫{(2x^2+1)/[x^2(x^2+1)]}dx的不定积分

∫{(2x^2+1)/[x^2(x^2+1)]}dx的不定积分

Answer 1

∫{(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。

Answer 2

∫{(2x^2+1)/[x^2(x^2+1)]}dx
=∫{[x^2+(x^2+1)]/[x^2(x^2+1)]}dx
=∫1/(x^2+1)]dx+∫1/x^2dx
=arctanx-1/x+c

Answer 3

Answer 4

到代换,用x=1/t,换元整理后挺简单的...