Finding a definte Integral

Question

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

Answer 1

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

 

A+B=1

A-B=0.

 

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.

Answer 2

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

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|