What is ∫ x^4-1 x^2 x^4+x^2+1 d x equal to?

Take option (1) $I_1=\sqrt{\frac{x^4+x^3+1}{x}}+C$$\begin{aligned} \frac{d I_1}{d x} & =\frac{d}{d x}\left[\left(x^3+x^2+x^{-1}\right)^{1 / 2}+C\right] \\ \frac{d I_2}{d x} & =\frac{1}{2}\left(x^3+x^2+x^{-1}\right)^{-1 / 2}\left(3 x^2+2 x-x^{-2}\right) \\ & =\frac{1}{2}\left[\frac{3 x^2+2 x-\frac{1}{x^2}}{\sqrt{x^3+x^2+\frac{1}{x}}}\right] \\ \frac{d I_2}{d x} & =\frac{1}{2}\left[\frac{3 x^4+2 x^3-1}{\frac{x^2}{\sqrt{x}} \sqrt{x^4+x^3+1}}\right] \end{aligned}$$$Take option (b) :$$\begin{aligned} I_2 & =\sqrt{x^4+2-\frac{1}{x^2}}+C \\ \frac{d I_2}{d x} & =\frac{1}{2}\left[x^4+2-x^{-2}\right]^{-1 / 2}\left[4 x^3+0+2 x^{-3}\right] \\ & =\frac{1}{2}\left[\frac{4 x^3+\frac{2}{x^3}}{\sqrt{x^4+2-\frac{1}{x^2}}}\right]=\frac{2 x^6+1}{\frac{x^3}{x} \sqrt{x^6+2 x^2-1}} \\ \frac{d I_2}{d x} & =\frac{2 x^6+1}{x^2 \sqrt{x^6+2 x-1}} \end{aligned}$$Take option (3) :$$\begin{aligned} & I_3=\sqrt{x^2+x^{-2}+1}+C \\ & \frac{d I_3}{d x}=\frac{1}{2}\left[x^2+x^{-2}+1\right]^{-1 / 2}\left[2 x-2 x^{-3}+0\right] \\ & =\frac{1}{2}\left[\frac{2 x-\frac{2}{x^3}}{\sqrt{x^2+\frac{1}{x^2}+1}}\right]=\frac{1}{2}\left[\frac{2\left(x^4-1\right)}{x^3 \sqrt{\frac{x^4+1+x^2}{x^2}}}\right] \\ & \frac{d I_3}{d x}=\frac{x^4-1}{x^2 \sqrt{x^4+x^2+1}} \end{aligned}$$\therefore$ Option (3) is correct answer.