∫1/x(x^2-1)^1/2dx

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∫1/x(x^2-1)^1/2dx
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∫1/x(x^2-1)^1/2dx
∫1/x(x^2-1)^1/2dx

∫1/x(x^2-1)^1/2dx
∫1/x(x^2-1)^1/2dx
u²=x²-1 2udu=2xdx x²=u²+1
∫1/x(x^2-1)^1/2dx
=∫1/{[u²+1]u}du
=∫(1/u)du-∫[1/(u²+1)]du
=lnu-arctanu+c
=ln(√x²-1)-arctan(√x²-1)+c

令x=sec t则 t=arccos(1/x)
∫1/x(x^2-1)^1/2dx=∫1/[sec t(sec^2 t-1)^1/2]d(sec t)=∫1/(sec t tan t)·sec t tan t dt==∫1dt=t+C=arccos(1/x)+C

∫ 1/[x√(x² - 1)] dx，令x = secz，dx = secztanz dz
= ∫ 1/[secz * |tanz|] * (secztanz dz)
= ∫ dz = ∫ dz
= arcsec(x) + C
= arccos(1/|x|) + C

∫[dx/(e^x(1+e^2x)]dx ∫1+2x/x(1+x)*dx∫1+2x/x(1+x) * dx 1/(x+x^2)dx ∫(x-1)^2dx, ∫x^1/2dx ∫1/x^2+x+1dx ∫1/(x^2+x+1)dx ∫dx/x^2(1-x^2) ∫dx/x^2(1+x^2) ∫X^2/1-x^2 dx. ∫X^2/1-x^2 dx. ∫ (x+arctanx)/(1+x^2) dx ∫（x^2+1/x^4)dx ∫2 -1|x²-x|dx ∫x^2/1+X dx. ∫ xe^x/(1+x)^2 dx ∫dx/[(1-x)x^2] ∫ x/(1+X^2)dx=