∫ ( ^ 2 -1) ( ^ 2 +1) ^ 4 +1 dx is equal to

$\mathrm{I}$=\int \frac{\mathrmx^{2}\left(1-\frac{1}{\mathrmx^{2}}\right) \mathrm{dx}}{\mathrmx^{2}\left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)\left(\mathrmx^{2}+\frac{1}{\mathrmx^{2}}\right)^{1 / 2}}$Let$\mathrm{x}+\frac{1}{\mathrm{x}}=\mathrm{p} \Rightarrow\left(1-\frac{1}{\mathrmx^{2}}\right) \mathrm{d} \mathrm{x}=\mathrm{dp}$\mathrm{I}$=\int \frac{\mathrm{dp}}{\mathrm{p} \sqrt{\mathrmp^{2}-2}}=\frac{1}{\sqrt{2}} \mathrm\sec^{-1} \frac{\mathrm{p}}{\sqrt{2}}=\frac{1}{\sqrt{2}} \sec ^{-1}$\left(\frac{\mathrmx^{2}+1}{\sqrt{2} \mathrm{x}}\right)+\mathrm{c}$