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Finding a definte Integral

Need help on     ∫1/(1-Χ²)    from -√2/2       to   +√2/2

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Ryan Y. | Harvey Mudd/Cornell grad; multi-subject, empathetic tutorHarvey Mudd/Cornell grad; multi-subject,...
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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.

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

1/(1-x2) = 1/[(1-x)(1+x)] 

1/(1-x2) = 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-x2) = A/(1-x)*[(1+x)/(1+x)] + B/(1+x)*[(1-x)/(1-x)]

1/(1-x2) = [A(1+x)+B(1-x)]/(1-x2)

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
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


= 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| 


Jonathan G. | A Tutor for the Upcoming Exam SeasonA Tutor for the Upcoming Exam Season
First note that we can use a partial fraction decomposition on the integrand:
1/(1-x2) = A/(1-x) + B/(1+x)
We now clear denominators and getting the resulting linear equations
This system has the solution A=B=1/2.
Thus our integral is 
∫ 1/(1-x2) = 1/2 ∫ 1/(1-x) + 1/2 ∫ 1/(1+x)
with the limits of integration being the original ones.  Let me know if you are unable to integrate the latter two integrals.  (HINT:  They are just natural logs.)  
For a slicker solution, note the original integrand is an even function in x and so we can change the original integral to twice the integral from 0 to root 2 over 2.