∫1/(2u^2(1-u^2))du等于多少?

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∫1/(2*u^2*(1-u^2))du等于多少?∫1/(2*u^2*(1-u^2))du等于多少?∫1/(2*u^2*(1-u^2))du等于多少?被积函数=1/[2u^2·(1-u^2)]=(1/

∫1/(2*u^2*(1-u^2))du等于多少?
∫1/(2*u^2*(1-u^2))du等于多少?

∫1/(2*u^2*(1-u^2))du等于多少?
被积函数 = 1/[2u^2·(1-u^2)] = (1/2)·[(1/u^2) + [1/(1 - u^2)]]
= (1/2)[u^(-2) + (1/2)[1/(1-u)] + (1/2)[1/(1+u)]]
= (1/2)u^(-2) + (1/4)[1/(1-u)] + (1/4)[1/(1+u)]
∴原积分 = (-1/2u) - (1/4)ln(1-u) + (1/4)ln(1+u) + C
=(-1/2u) + (1/4)ln[(1+u)/(1-u)] + C

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