∫ln(x+√(x^2-1)dx,

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∫ln(x+√(x^2-1)dx,
∫ln(x+√(x^2-1)dx,

∫ln(x+√(x^2-1)dx,
很高兴为您解答,解题步骤如下,

或者如果不要过程,我们可以：

令x=cosa，则x+√(x^2-1=√2sin(a+45)/2
接下来就会了吧

令y=ln(x+√(x^2-1))
=>e^y = x+√(x^2-1)
已知ln(x+√(x^2-1))+ln(x-√(x^2-1))=ln(xx-xx+1)=0
=>
-y = ln(x-√(x^2-1))
=>
e^-y = x-√(x^2-1)
=>
x = 1/2*(e^y+e^-y)
=>
∫ln(x+√...

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令y=ln(x+√(x^2-1))
=>e^y = x+√(x^2-1)
已知ln(x+√(x^2-1))+ln(x-√(x^2-1))=ln(xx-xx+1)=0
=>
-y = ln(x-√(x^2-1))
=>
e^-y = x-√(x^2-1)
=>
x = 1/2*(e^y+e^-y)
=>
∫ln(x+√(x^2-1)dx=∫ydx=1/2*(∫y(e^y-e^-y)dy
已知：
∫ye^ydy=∫yd(e^y) = ye^y-∫e^ydy=e^y(y-1)+C
-∫ye^-ydy=-∫ye^ydy=-e^y(y-1)+C
=>
∫ln(x+√(x^2-1)dx=C

收起

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