高数求不定积分
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令(x+1)^1/3=t,x=t^3-1,dx=3t^2dt
∫dx/[1+(x+1)^1/3]
=∫3t^2/(1+t)dt
=3∫t^2/(1+t)dt
=3∫(t^2-1+1)/(1+t)dt
=3∫[t-1+1/(1+t)]dt
=3/2t^2-3t+3ln(1+t)+C
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
∫dx/[1+(x+1)^1/3]
=∫3t^2/(1+t)dt
=3∫t^2/(1+t)dt
=3∫(t^2-1+1)/(1+t)dt
=3∫[t-1+1/(1+t)]dt
=3/2t^2-3t+3ln(1+t)+C
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
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