\int \frac{(1-\frac{1}{x^2})dx}{(\sqrt{x+\frac{1}{x}+\alpha})(\sqrt{x+\frac{1}{x}+\beta})}
put x+\frac{1}{x}=t and u r done
answer is
\ln{\left( \frac{2(\sqrt{x^2+\alpha x+1)(x^2+\beta x +1)}}{x}+\alpha+\beta+2x\right)}+K
(2) $Calculate $\int\frac{(x^2-1)}{x.\sqrt{x^2+\alpha x+1}.\sqrt{x^2+\beta x+1}}dx$
\int \frac{(1-\frac{1}{x^2})dx}{(\sqrt{x+\frac{1}{x}+\alpha})(\sqrt{x+\frac{1}{x}+\beta})}
put x+\frac{1}{x}=t and u r done
answer is
\ln{\left( \frac{2(\sqrt{x^2+\alpha x+1)(x^2+\beta x +1)}}{x}+\alpha+\beta+2x\right)}+K
I ' ll outline a proof - you will be able to solve then I hope -
Take out an " x " each from the two square roots . You will be left with a " x 2 " in the denominator and the two square roots . Then , divide the numerator by " x 2 " . Here , write " x + 1x = z " . You will get " dz " on the numerator only , and you ' ll also be left with a standard integral .