Integral....

on seeing the limits , the natural instinct is to put 1x=t
and then proceed

$Extremely Sorry for posting a Wrong Integral.....\\\\ The Righta Integral is $\int_{\frac{1}{2}}^{2}\frac{1}{(3x^2-10x+3)(x^{2010}+1)}dx$

I=\int_{\frac{1}{2}}^{2}{\frac{dx}{(3x-1)(x-3)(1+x^{2010})}}
t=\frac{1}{x}

now u get

I=\int_{\frac{1}{2}}^{2}{\frac{t^{2010}dt}{(3t-1)(t-3)(1+t^{2010})}}

2I=\int_{\frac{1}{2}}^{2}{\frac{dx}{(3x-1)(x-3)}}

this can be done easily

Ans=-\frac{\ln(25)}{16}