根号下x^2-1原函数

2024年11月14日 13:31
有1个网友回答
网友(1):

∵∫√(x^2-1)dx
=x√(x^2-1)-∫xd[√(x^2-1)]
=x√(x^2-1)-∫x[x/√(x^2-1)]dx
=x√(x^2-1)-∫[(x^2-1+1)/√(x^2-1)]dx
=x√(x^2-1)-∫√(x^2-1)dx-∫[1/√(x^2-1)]dx,

∴2∫√(x^2-1)dx
=x√(x^2-1)-∫[1/√(x^2-1)]dx
=x√(x^2-1)-∫{[x+√(x^2-1)][1/√(x^2-1)]/[x+√(x^2-1)]}dx
=x√(x^2-1)-∫{[x/√(x^2-1)+1]/[x+√(x^2-1)]}dx
=x√(x^2-1)-∫{1/[x+√(x^2-1)]}d{[x+√(x^2-1)]}
=x√(x^2-1)-ln[x+√(x^2-1)]+C,
∴∫√(x^2-1)dx=(1/2)x√(x^2-1)-(1/2)ln[x+√(x^2-1)]+C。