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∫1/(x4(√x2%1)) Dx

解: 令x=secu,则dx=secu·tanu du ∫1/[x^4·√(x^2-1)]dx =∫(cosu)^3 du =∫[1-(sinu)^2]cosu du =∫[cosu-(sinu)^2·cosu] du =∫cosu du-∫(sinu)^2·cosudu =sinu-∫(sinu)^2 d(sinu) =sinu-1/3·(sinu)^3+C =√(1-1/x^2)-1/3·(1-1/x^2)^(3/2)+C

解:令x2=u ∫[x/(1+x?)]dx =?∫[1/(1+x?)]d(x2) =?∫[1/(1+u2)]du =?arctanu +C =?arctan(x2) +C

∫ 1/[x^4(x²+1)] dx =∫ (1+x²-x²)/[x^4(x²+1)] dx =∫ (1+x²)/[x^4(x²+1)] dx - ∫ x²/[x^4(x²+1)] dx =∫ 1/x^4 dx - ∫ 1/[x²(x²+1)] dx =-1/(3x³) - ∫ 1/x² dx + ∫ 1/(x²+1) d...

如图所示:

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...

∫(x2+1)/(x4+1)dx=∫(1+1/x^2)/(x^2+1/x^2)dx=∫{1/[2+(x-1/x)^2]}d(x-1/x) =(√2/2)arctan[(x-1/x)/√2]+c(希望能帮到你,麻烦在我回答的下面点击 “好评”,谢谢你啦^_^)

令x=tany ∫(x^2/(1+x^4))dx =∫(tany^2/(1+tany^4))*(1/(cosy)^2)dy =∫(siny)^2/((siny)^4+(cosy)^4) dy =∫(1/2)(1-cos2y)/(1-4(siny)^2(cosy)^2) dy =(1/2)∫(1-cos2y)/(1-(sin2y)^2) dy =(1/2)∫1/(1-(sin2y)^2) dy - (1/2)∫cos2y/(1-(sin2y)^2) ...

1-x4分之1的积分 =1/2 ∫[1/(1-x²)+1/(1+x²)]dx =-1/2∫1/(x²-1)dx+1/2 ∫1/(1+x²)dx =-1/4ln|(x-1)/(x+1)|+1/2arctanx+c

∫ (1 + x²) / (1 + x^4) dx = ∫ (1 + 1 / x²) / (x² + 1 / x²) dx = ∫ d(x - 1 / x) / [(x - 1 / x)² + 2] = (1 / √2) arctan[(x - 1 / x) / √2] + C

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