Hi Fermin,

I confess, this one had me stumped for a bit, so I had to use Wolfram Alpha to calculate a solution. Once I found the solution, I was able to figure out how to get there.

I confess, this one had me stumped for a bit, so I had to use Wolfram Alpha to calculate a solution. Once I found the solution, I was able to figure out how to get there.

What you need to do is use partial fractions to break up the integrand.

1/(1-x

1/(1-x

1/(1-x

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

^{2}) = A/(1-x) + B/(1+x).where A and B are unknown. You find A and B by getting everything on the right side under a common denominator and making sure the two sides are equal.

1/(1-x

1/(1-x

Because everything on the right side has to equal everything on the left side, and because both sides have the same denominator,

1 = A(1+x) + B(1-x)

A(1) + B(1) = 1

A(x) + B(-x) = 0

A = B

1/(1-x

^{2}) = A/(1-x)*[(1+x)/(1+x)] + B/(1+x)*[(1-x)/(1-x)]1/(1-x

^{2}) = [A(1+x)+B(1-x)]/(1-x^{2})Because everything on the right side has to equal everything on the left side, and because both sides have the same denominator,

1 = A(1+x) + B(1-x)

A(1) + B(1) = 1

A(x) + B(-x) = 0

A = B

A + A = 1

A = 1/2

B = 1/2

How does this help us? Well, by decomposing the integrand, we now have

B = 1/2

How does this help us? Well, by decomposing the integrand, we now have

int[ 1/[2(x+1)] + 1/[2(x-1)] ]dx

which can be broken into two separate integrals:

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

and evaluated

=

or

which can be broken into two separate integrals:

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

and evaluated

=

**(1/2)ln |x+1| -(1/2)ln|x-1| +C**or

= tanh

^{-1 }x + C. (the inverse hyperbolic tangent function)

Now, because it's a definite integral, we actually have to evaluate it (and drop the constant of integration).

= (1/2)ln |√2/2+1| -(1/2)ln|√2/2-1| - [(1/2)ln |-√2/2+1| -(1/2)ln|-√2/2-1)|

= (1/2)ln |√2/2+1| -(1/2)ln|√2/2-1| - [(1/2)ln |√2/2-1| -(1/2)ln|√2/2+1)|

=

**ln |√2/2+1| -ln|√2/2-1|**

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