Question

不定积分∫(1/x^2)√(1-x)/(1+x)dx如何解？

还有一道，∫1/(1+x^3)dx，求解

Answer 1

∫√[(1-x)/(1+x)] dx/x^2

x=cosu dx=sinu √[(1-x)/(1+x)]=(1-cosu)/sinu

原式=∫(1-cosu)du/(cosu)^2

=∫du/(cosu)^2-∫cosudu/(1-sinu)(1+sinu)

=tanu-∫dsinu/[(1-sinu)(1+sinu)]

=tanu-ln[|1+sinu|/|cosu|] +C

=√[(1/x^2)-1 ] -ln[|1+√(1-x^2)|/|x| ]+C

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

x^2/x(1+x^2)化简得x/(1+x^2)
把分母x收到dx中为d(1+x^2)
再乘以系数1/2
就好算了
或
x+(1/2)x^2+(1/2)ln(1+x^2)+C,C为常数，）原式＝[(x^2+1)+x]/x(1+x^2)=1/x+1/(1+x^2),积分为lnx+arctanx+C,C为常数

Answer 3

∫√[(1-x)/(1+x)] dx/x^2
x=cosu dx=sinu √[(1-x)/(1+x)]=(1-cosu)/sinu
原式=∫(1-cosu)du/(cosu)^2
=∫du/(cosu)^2-∫cosudu/(1-sinu)(1+sinu)
=tanu-∫dsinu/[(1-sinu)(1+sinu)]
=tanu-ln[|1+sinu|/|cosu|] +C
=√[(1/x^2)-1 ] -ln[|1+√(1-x^2)|/|x| ]+C

∫dx/(1+x^3)
=∫dx/[(1+x)(1-x+x^2)]=(1/3)∫(1+x)dx/[x(1-x+x^2)]-(1/3)∫dx/[x(1+x)]
=(1/3)∫xdx/[x^2(1-x+x^2)]+(1/3)∫dx/[(x-1/2)^2+3/4]-(1/3)ln[|1+x|/|x|]
=(1/6)∫dx^2/(x^2(1-x+x^2)+... ...=(2/3√3)arctan(2x/√3-1/√3)-(1/3)ln|1+x|/|x|
=(1/6)∫dx^2/[(x-1)(1-x+x^2]+(1/6)∫dx^2/[(x-1)x^2]+...
=(1/3)∫xdx/[(x-1)(x^2-x+1)]+(1/3)∫dx/[x(x-1)]+...
=(1/3)∫(x-1)dx/(x^2-x+1)-(1/3)∫dx/(x-1)+(1/3)∫dx/(x-1)-(1/3)∫dx/x+...
=(1/6)ln|x^2-x+1|-(1/6)∫dx/[(x-1/2)^2+3/4]-(1/3)ln|x|+...

=(1/6)ln|x^2-x+1|+(1/3√3)arctan(2x/√3-1/√3)-(1/3)ln|1+x|+C

Answer 4

x^8=(x^8-1)+1=(x^4+1)(x^2+1)(x^2-1)+1
所以x^8/(1+x^2)=(x^4+1)(x^2-1)+1/(1+x^2)
这两部分的积分相信你肯定没问题的

Answer 5

令x=cos2a 则∫(1/x^2)√(1-x)/(1+x)dx=∫(1/cos^2(2a)*根号((1-cos2a)/(1+cos2a))dcos2a
=∫(1/cos^2(2a)*根号(2sin^2(a)/(2cos^2(a))dcos2a=∫(1/cos^2(2a)*sina/cosa dcos2a
=∫(sina/(cos^2(2a)*cosa) dcos2a
=∫sina/(cos^2(2a)*cosa) *(-2sin2a) da
=-4∫(sin^2(a)/(cos^2(2a)) da
= -4∫1/(csc^a-2) da
=-4∫1/(cot^2(a)-1) da