How to integrate 1/x^2 (1 - x^2)

[teng125]

May i know how to integ 1/x^2 (1-x^2) ??
i try to form Ax/x^2 + Bx+C/(1-x^2) but can't get the answer...


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

 

[fargoth]

use latex already!
anyway, the form should be $$\frac{Ax+B}{x^2}+\frac{Cx+D}{1-x^2}$$
its alway a polynom with one order lower...

 

[VietDao29]

use latex already!
anyway, the form should be $$\frac{Ax+B}{x^2}+\frac{Cx+D}{1-x^2}$$
its alway a polynom with one order lower...

In fact, you should notice that 1 - x² = (1 - x) (1 + x).
So partial fraction it gives:
$$\frac{1}{x ^ 2 (1 - x ^ 2)} = \frac{A}{x} + \frac{B}{x ^ 2} + \frac{C}{1 - x} + \frac{D}{1 + x}$$.
There are several ways to do this:
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The most common way is to multiply both sides by x²(1 - x²), you'll have:
$$1 = Ax(1 - x ^ 2) + B(1 - x ^ 2) + Cx ^ 2 (1 + x) + Dx ^ 2 (1 - x)$$.
Now plug some value of x in the RHS of the equation and solvwe for x. You'll need to plug in at least 4 values of x, since there are 4 unknowns (A, B, C, and D). You can choose the value of x such that the RHS become as simple as possible. For example, you can choose x = 0 (A, C, and D will disappear in the RHS), x = 1 (A, B, D will disappear), x = -1 (A, B, C will disappear), and choose a random x (since you'll need 4 values of x) (e.g: x = 2).
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The second way is to notice that the LHS of this equation:
$$\frac{1}{x ^ 2 (1 - x ^ 2)} = \frac{A}{x} + \frac{B}{x ^ 2} + \frac{C}{1 - x} + \frac{D}{1 + x}$$
is an even function, so the RHS must also be an even function.
That means:
$$\frac{1}{x ^ 2 (1 - x ^ 2)} = \frac{A}{x} + \frac{B}{x ^ 2} + \frac{C}{1 - x} + \frac{D}{1 + x} = - \frac{A}{x} + \frac{B}{x ^ 2} + \frac{C}{1 + x} + \frac{D}{1 - x}$$.
That means A = 0, C = D, so you are left with 2 unknowns:
$$\frac{1}{x ^ 2 (1 - x ^ 2)} = \frac{B}{x ^ 2} + \frac{C}{1 - x} + \frac{C}{1 + x}$$.Plug in 2 values of x, and solve for B, and C.
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The third way (even faster) is to notice the fact that:
$$\frac{1}{x ^ 2} + \frac{1}{1 - x ^ 2} = \frac{1 - x ^ 2 + x ^ 2}{x ^ 2 (1 - x ^ 2)} = \frac{1}{x ^ 2 (1 - x ^ 2)}$$.
So:
$$\frac{1}{x ^ 2 (1 - x ^ 2)} = \frac{1}{x ^ 2} + \frac{1}{1 - x ^ 2}$$.
Just do the same for $$\frac{1}{(1 - x ^ 2)} = \frac{1}{(1 - x) (1 + x)}$$. Can you go from here?