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