Integrate 1/ x^2 (1-x) (1-x^2)?

[teng125]

how to integ 1/ x^2 (1-x) (1-x^2)??

the answer is
-1/x + (1/2) ln (1+x) -(1/2) ln (1-x)

 

[fargoth]

please learn to use latex, did you mean to integrate this: $$\frac{1}{x^2(1-x)(1-x^2)}$$?

(click on it to see how i wrote it...)

 

[teng125]

ya,this is what i mean

 

[fargoth]

try splitting the exprassion: $$\frac{1}{x^2(1-x)^2(1+x)}=\frac{1}{x^2(1-x)^2}-\frac{1}{x(1-x)^2(1+x)}$$
for example... you can keep simplifying it.

 

[fargoth]

its easy to get rid of x (put 1+x-x), the generic way to simplify is $$\frac{1}{(1+x)(1-x^2)}=\frac{A}{1+x}+\frac{Bx+C}{1-x^2}$$
and A B and C for them to sattisfy the equation.

 

[Tx]

Simplify the bottom to,
1/ x^2(x-1)^2(x+1)
Then:
1 = Ax/x^2 + Cx+D/(x-1)^2 + E/(x+1)
1= Ax(x-1)^2(x+1) + (Cx+D)x^2(x+1) + E(x-1)^2x^2

Collect like terms and solve for A -> E and then integrate.

 

[teng125]

oh...okok thanks very much