Question

根号下（1+u^2)+u=x 求解u

这个怎么求，书上的答案是直接写出了根号下（1+u^2)-u=1/x ，这个式子是怎么得来的?

Answer 1

平方差公式：
[根号下（1+u^2)+u][根号下（1+u^2)-u] = 1
x[根号下（1+u^2)-u] = 1
[根号下（1+u^2)-u] = 1/x

Answer 2

√(1+u^2)+u=x

[√(1+u^2)+u]*[√(1+u^2)-u]=x[√(1+u^2)-u]
(1+u^2)-u^2=x[√(1+u^2)-u]
即1=x[√(1+u^2)-u]
推出√(1+u^2)-u=1/x

Answer 3

左右两边倒数一下吧！

Answer 4

∫ (1+u^2)/(1-u^2)d u
= ∫ (-1+ 2/(1-u^2)) du
= -u + 2∫ (1/(1-u^2)) du
= -u + ∫ (1/(1-u) + 1/(1+u)) du
= -u -ln|1-u| +ln|1+u| +C