∫ln(x+√(x^2-1)dx,

来源：学生作业帮助网编辑：作业帮时间：2024/11/15 07:48:16

∫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+√...

全部展开

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

求不定积分∫ln(x+√(x^2+1))dx 求不定积分：∫ ln(x+√(1+x^2) )dx 定积分∫ ln(√1+x^2+x)dx 不定积分：∫ ln(x+√1+x^2) dx ∫ln(x+√(x^2-1)dx, ∫ln(x+√(1+x^2))dx 求不定积分 ∫ln(x+√(1+x^2))dx 求不定积分求不定积分∫dx/x√1-ln^2 x 是ln平方的x ∫ln(1+√x)dx ∫ln(1+x^2)dx ∫ ln(x^2 -1)dx 步骤 ∫ln(x/2)dx ∫(ln ln x + 1/ln x)dx ∫ln/x√(1-ln^2x)dx不定积分求解 ∫1+x^2 ln^2x / x lnx dx ∫x*ln(x²+1)dx ∫x*ln(x-1)dx ∫x* ln (x-1) dx ∫(ln(x+2)-ln(x+1))/(x^2+3x+2)dx=