\(\int x \sqrt{\frac{2 \sin \left(x^2+1\right)-\sin 2\left(x^2+1\right)}{2 \sin \left(x^2+1\right)+\sin 2\left(x^2+1\right)}} \mathrm{d} x=\)

\(\begin{aligned} & \text { Let } \mathrm{I}=\int x \sqrt{\frac{2 \sin \left(x^2+1\right)-\sin 2\left(x^2+1\right)}{2 \sin \left(x^2+1\right)+\sin 2\left(x^2+1\right)}} \mathrm{d} x \\ & =\int x \sqrt{\frac{2 \sin \left(x^2+1\right)-2 \sin \left(x^2+1\right) \cos \left(x^2+1\right)}{2 \sin x\left(x^2+1\right)+2 \sin \left(x^2+1\right) \cos \left(x^2+1\right)}} \mathrm{d} x \\ & =\int x \sqrt{\frac{1-\cos \left(x^2+1\right)}{1+\cos \left(x^2+1\right)}} d x \\ & =\int x \sqrt{\frac{2 \sin ^2\left(\frac{x^2+1}{2}\right)}{2 \cos ^2\left(\frac{x^2+1}{2}\right)}} d x \\ & =\int x \tan \left(\frac{x^2+1}{2}\right) \mathrm{d} x \\ & \text { Let }\left(\frac{x^2+1}{2}\right)=\mathrm{t} \Rightarrow x \mathrm{~d} x=\mathrm{dt} \\ & \therefore \quad \mathrm{I}=\int \tan \mathrm{t} d \mathrm{t} \\ & =\log (\sec \mathrm{t})+\mathrm{c} \\ & =\log \left(\sec \left(\frac{x^2+1}{2}\right)\right)+\mathrm{c} \\ & \end{aligned}\)