1/( x^1/2+x^1/4)的不定积分

xzyzdana2022-10-04 11:39:541条回答

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kill73174 共回答了20个问题 | 采纳率90%: 45 令 x^(1/4)=u,则 x=u^4,dx=4u^3du
∫dx/[x^(1/2)+x^(1/4)] = ∫4u^3du/[u^2+u]
= 4∫[u-1+1/(u+1)]du = 2u^2-4u+4ln|u+1|+C
= 2x^(1/2)-4x^(1/4)+4ln[x^(1/4)+1]+C; 1年前

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求不定积分,分子是1,分母是x^1/2+x^1/4 明月何曾照两乡1年前1: tbbh0072 共回答了18个问题 | 采纳率88.9%

设x^1/4=t
x^1/2=t^2
x=t^4
dx=4t^3dt
∫dx/(x^1/2+x^1/4 )= ∫4t^3dt/(t^2+t)
= ∫4t^2dt/(t+1)
=4∫(t^2-1+1)dt/(t+1)
=4∫(t-1)dt+4∫dt/(t+1)
=4*(t^2/2-t)+4ln|t+1|+C
=2t^2-4t+4ln(t+1)+C
=2x^1/2-x^1/4+4ln(x^1/4+1)+C

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