√(X^2-1)的不定积分

Question

一楼的怎么求 √(x²-1) / x 的不定积分了 ？
二楼的打错了吧 看不懂。。

Answer 1

设x=sect

√(x²-1)=tant

d(sect)=sec(t)*tan(t)dt

∫[√(x²-1) ]dx

=∫tan(t)sec(t)*tan(t)dt

=∫sin^2(t)/cos^3(t)tdt

=∫[1-cos^2(t)]/cos^3(t)dt

=∫1/cos^3(t)dt - ∫1/costdt

=(x*(x^2 - 1)^(1/2))/2 - log(x + (x^2 - 1)^(1/2))/2

不定积分的公式：

1、∫ a dx = ax + C，a和C都是常数

2、∫ x^a dx = [x^(a + 1)]/(a + 1) + C，其中a为常数且 a ≠ -1

3、∫ 1/x dx = ln|x| + C

4、∫ a^x dx = (1/lna)a^x + C，其中a > 0 且 a ≠ 1

5、∫ e^x dx = e^x + C

6、∫ cosx dx = sinx + C

7、∫ sinx dx = - cosx + C

8、∫ cotx dx = ln|sinx| + C = - ln|cscx| + C

Answer 2

一楼是正确的：
∫√（x^2-1）dx =∫tanx * secx*tanxdx (第二类换元法：x=sect,t属于<0,π/2))
=∫sect(sect*sect-1)dt=∫sect*sect*sectdt-∫sectdt=∫sectdtant-∫sectdt
=secttant-∫tant*tant*sectdt-∫sectdt
即∫√（x^2-1）dx =∫tant * sect*tantdt= secttant-∫tant*tant*sectdt-∫sectdt
将等式右边的∫tant*tant*sectdt移到左边:
∫√（x^2-1）dx =∫tant * sect*tantdt=1/2 secttant-1/2∫sectdt
=1/2 secttant-1/2ln⁄sect+tant⁄+c=1/2x√(x^2-1)-1/2ln(x+√(x^2-1))+c本回答被提问者采纳

Answer 3

不定积分=1/2*（x^2-1）^(-1/2) *2x+c
=x/√(x^2-1)+c (c为常数)
搞错啦
不好意思

Answer 4

设x=sect
√(x²-1)=tant
d(sect)=sec(t)*tan(t)dt

∫[√(x²-1) ]dx
=∫tan(t)sec(t)*tan(t)dt
=∫sin^2(t)/cos^3(t)tdt
=∫[1-cos^2(t)]/cos^3(t)dt
=∫1/cos^3(t)dt - ∫1/costdt
=(x*(x^2 - 1)^(1/2))/2 - log(x + (x^2 - 1)^(1/2))/2本回答被网友采纳

Answer 5

利用双曲代换令t=chx