Integrate the following w.r.t x

∫ (x^2-1)/(x^4-x^2-1)dx

Let, I = ∫ (x^2-1)/(x^4-x^2-1)dx

= ∫ (x^2-1)/x^2(x^2-1+1/x^2)dx

= ∫ (1-1/x^2)/[(x+1/x^2)-3]

say x+1/x=z => (1-1/x^2)dx=dz


So, I= ∫ dz/[z^2-(√3)^2]

= 1/2√3.log|(z-√3)/(z+√3)|+c

= 1/2√3.log|(x^2- √3x+1)/( x^2+√3x+1)|+C Ans.

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