How do you use partial fraction decomposition to decompose the fraction to integrate #(x^2+x+1)/(1-x^2)#?

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Answer 1 · 2015-07-27T16:24:09

Divide first, then find the partial fraction decomposition.
#(x^2+x+1)/(1-x^2) = -1 - (3/2)/(x-1) + (1/2)/(x+1)#

Before looking for a partial fraction decomposition, we must have the degree of the denominator strictly less than that of the numerator.

So we need to divide or regroup to get:

#(x^2+x+1)/(1-x^2) = (x^2-1+x+2)/(1-x^2)#

# = (x^2-1)/(1-x^2)+(x+2)/(1-x^2)#

# = -1 - (x+2)/(x^2-1)#

Now we can get the partial fraction decomposition for:

#(x+2)/(x^2-1) = (x+2)/((x-1)(x+1)) = A/(x-1)+B/(x+1)#

We need #Ax+A+Bx-B = x+2#

So

#A+B = 1#
#A-B = 2#

#2A = 3# so #A = 3/2# and #B = -1/2#

Putting it all together we have:

#(x^2+x+1)/(1-x^2) = -1-((3/2)/(x-1) - (1/2)/(x+1))#

# = -1 - (3/2)/(x-1) + (1/2)/(x+1)#

Now integrate.