How do you integrate #int (x^2+x+1)/(1-x^2)# using partial fractions?

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Answer 1 · 2017-08-02T17:03:58

The answer is #=-x+1/2ln(|x+1|)-3/2ln(|x-1|)+C#

Let's simplify the quotient

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

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

Now, we perform the decomposition into partial fractions

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

The numerators are the same. we compare the numerators

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

Let #x=1#, #=>#, #3=2B#, #=>#, #B=3/2#

Let #x=-1#, #=>#, #1=-2A#, #=>#, #A=-1/2#

Therefore,

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

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

#=-x+1/2ln(|x+1|)-3/2ln(|x-1|)+C#