1thanks asish , after getting f(x) = x2-x-3
∫f(x)/x3-1
=∫x3-x-3/(x-1)(x2+x+1)
by partial fractions
x3-x-3/(x-1)(x2+x+1)= A/(x-1) + Bx+c/(x2+x+x)
A=-1 , B =2 , C =2
∫-1dx/x-1 + ∫ (2x+2)/x2+x+1
=-log|x-1|+log|x2+x+1| +2/√3tan-12x+1/√3
∫ f(x)/x3-1 where f(x) is a polynomial of degree 2 in x such that f(0)=f(1)=3f(2)=-3
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4 Answers
1arey yaar amit tu toh integration sums ke peeche haath dho kar padh gaya hai .. ::))
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