∫根号下(1+x^1/2)dx

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∫根号下(1+x^1/2)dx∫根号下(1+x^1/2)dx∫根号下(1+x^1/2)dx令a=√(1+√x)x=(a²-1)²=a^4-2a²+1dx=(4a³

∫根号下(1+x^1/2)dx
∫根号下(1+x^1/2)dx

∫根号下(1+x^1/2)dx

令a=√(1+√x)
x=(a²-1)²=a^4-2a²+1
dx=(4a³-4a)da
所以原式=∫a(4a³-4a)da
=∫(4a^4-4a²)da
=4a^5/5-4a³/3+C
=4(1+√x)²√(1+√x)/5-4(1+√x)√(1+√x)/3+C

∫√(1+√x) dx
let
√x = (tany)^2
[1/(2√x)]dx = 2tany. (secy)^2 dy
dx = 4(tany)^3. (secy)^2 dy
∫√(1+√x) dx
=4∫(tany)^3. (secy)^3 dy
=4∫ (tany)^2 (secy)^2 dsecy
=4∫ [(secy)^...

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∫√(1+√x) dx
let
√x = (tany)^2
[1/(2√x)]dx = 2tany. (secy)^2 dy
dx = 4(tany)^3. (secy)^2 dy
∫√(1+√x) dx
=4∫(tany)^3. (secy)^3 dy
=4∫ (tany)^2 (secy)^2 dsecy
=4∫ [(secy)^4 - (secy)^2] dsecy
= 4[ (secy)^5/5 - (secy)^3/3 ]+ C
= 4[ (√(1+√x))^5/5 - (√(1+√x))^3/3 ]+ C
where
√x = (tany)^2
x^(1/4) = tany
secy = √(1+√x)

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