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Answer» 1/(x - x^3) = 1/[x(1 - x^2)] and 1/[x(1 - x^2)] = A/x + (Cx + B)/(1 - x^2) multiplying both sides by x(1 - x^2) gives 1 = A(1 - x^2) + (Cx + B)x For x=0 you have 1 = A -----> so A=1 For x=1 you have 1 = (C + B) For x=-1 you have 1 = (-C + B)-11 = C - B The system is C + B = 1C - B = 1 Adding these two equations together gives 2C = 2 -------> so C = 1 Back substituting shows that B=0. This means that 1/[x(1 - x^2)] = 1/x + x/(1 - x^2) Substituting into the original equation gives dy=[1/x + x/(1 - x^2) ] dx Integrating both sides gives y = ln\|x\| - (1/2)ln\|1 - x^2\| + C Since 1-x^2 = (1 - x)(1+x), this could also be written as y = ln\|x\| - (1/2)ln\| (1 - x)(1+x) \| + C = ln\|x\| - (1/2)ln\|1 - x\| - (1/2)ln\|1 + x\| + C

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Answer»

1/(x - x^3) = 1/[x(1 - x^2)]

and

1/[x(1 - x^2)] = A/x + (Cx + B)/(1 - x^2)

multiplying both sides by x(1 - x^2) gives

1 = A(1 - x^2) + (Cx + B)x

For x=0 you have

1 = A -----> so A=1

For x=1 you have

1 = (C + B)

For x=-1 you have

1 = (-C + B)-11 = C - B

The system is

C + B = 1C - B = 1

Adding these two equations together gives

2C = 2 -------> so C = 1

Back substituting shows that B=0. This means that

1/[x(1 - x^2)] = 1/x + x/(1 - x^2)

Substituting into the original equation gives

dy=[1/x + x/(1 - x^2) ] dx

Integrating both sides gives

y = ln|x| - (1/2)ln|1 - x^2| + C

Since 1-x^2 = (1 - x)(1+x), this could also be written as

y = ln|x| - (1/2)ln| (1 - x)(1+x) | + C

= ln|x| - (1/2)ln|1 - x| - (1/2)ln|1 + x| + C