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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

∫ 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...

∫(x2+1)/(x4+1)dx=∫(1+1/x2)/(x2+1/x2)dx=∫1/[(x-(1/x))^2+2]d(x-(1/x))=[(√2)/2]arctan{[x-(1/x)]/√2}+C

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...

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) ...

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原式=1/(x^2-1)(x^2+1)=1/2*(1/(x^-1)-1/(x^2+1))=1/4(1/(1-x)-1/(1+x))-1/2*(1/(x^2+1)) 对上式积分得1/4*(ln(x-1)-ln(x+1))+1/2*acrtanx

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