Question

x/（1+x^4）^2的积分

Answer 1

简单计算一下即可，答案如图所示

Answer 2

你好！∫x^4/(1+x^2)dx=∫(x^4-1+1)/(1+x^2)dx=∫(x^2-1+1/(1+x^2))dx=(1/3)x^3-x+arctanx+c。经济数学团队帮你解答，请及时采纳。谢谢！

Answer 3

∫xdx/(1+x^4)^2 = (1/2)∫d(x^2)/(1+x^4)^2 (令 x^2 = tanu)
= (1/2)∫(secu)^2du/(secu)^4 = (1/2)∫(cosu)^2du
= (1/4)∫(1+cos2u)du = (1/4)[u+(1/2)sin2u] + C
= (1/4) [arctan(x^2) + x^2/(1+x^4)] + C
验算：(1/4)[arctan(x^2) + x^2/(1+x^4)]'
= (1/4){2x/(1+x^4)+[2x(1+x^4)-4x^5]/(1+x^4)^2]}
= (x/2)[1/(1+x^4)+(1-x^4)/(1+x^4)^2]
= x/(1+x^4)^2

Answer 4

∫[1/(4-x2)]dx =?∫[1/(x+2) -1/(x-2)]dx =?[ln|x+2|-ln|x-2|] +C =?ln|(x+2)/(x-2)| +C

Answer 5

一楼的结果不对，那个结果求导后为：x²/(x²-1)²

本题解法技巧较高
∫ x²/(1+x^4) dx
=(1/2)∫ (x²-1+x²+1)/(1+x^4) dx
=(1/2)∫ (x²-1)/(1+x^4) dx + (1/2)∫ (x²+1)/(1+x^4) dx
分子分同除以x²
=(1/2)∫ (1-1/x²)/(1/x²+x²) dx + (1/2)∫ (1+1/x²)/(1/x²+x²) dx
分子放到微分之后
=(1/2)∫ 1/(1/x²+x²) d(x+1/x) + (1/2)∫ 1/(1/x²+x²) d(x-1/x)
=(1/2)∫ 1/(1/x²+x²+2-2) d(x+1/x) + (1/2)∫ 1/(1/x²+x²-2+2) d(x-1/x)
=(1/2)∫ 1/[(x+1/x)²-2] d(x+1/x) + (1/2)∫ 1/[(x-1/x)²+2] d(x-1/x)
=(√2/8)ln|(x+1/x-√2)/(x+1/x+√2)| + (√2/4)arctan[(x-1/x)/√2] + C
=(√2/8)ln|(x²+1-√2x)/(x²+1+√2x)| + (√2/4)arctan[(x-1/x)/√2] + C

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