JEE Main 2021 — Mathematics Calculus

We have,

$${\int }_{\frac{-1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}}{({(\frac{x+1}{x-1})}^{2}+{(\frac{x-1}{x+1})}^{2}-2)}^{\frac{1}{2}}dx$$∫2​−1​2​1​​((x−1x+1​)2+(x+1x−1​)2−2)21​dx

$$={\int }_{-\frac{1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}}\sqrt{{[\frac{x-1}{x+1}-\frac{x+1}{x-1}]}^{2}}dx$$=∫−2​1​2​1​​[x+1x−1​−x−1x+1​]2​dx

$$={\int }_{\frac{-1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}}\sqrt{{(\frac{-4x}{{x}^{2}-1})}^{2}}dx$$=∫2​−1​2​1​​(x2−1−4x​)2​dx

$$={\int }_{\frac{-1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}}\sqrt{{(\frac{4x}{1-{x}^{2}})}^{2}}dx$$=∫2​−1​2​1​​(1−x24x​)2​dx

$$={\int }_{\frac{-1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}}\sqrt{\frac{16{x}^{2}}{{(1-{x}^{2})}^{2}}}dx$$=∫2​−1​2​1​​(1−x2)216x2​​dx

$$={\int }_{\frac{-1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}}{\frac{4|x|}{(1-{x}^{2})}}dx$$=∫2​−1​2​1​​(1−x2)4∣x∣​dx

$$=2{\int }_{0}^{\frac{1}{\sqrt{2}}}{\frac{4|x|}{(1-{x}^{2})}}dx$$=2∫02​1​​(1−x2)4∣x∣​dx

$$=4{\int }_{0}^{\frac{1}{\sqrt{2}}}(\frac{2x}{1-{x}^{2}})dx$$=4∫02​1​​(1−x22x​)dx

$$=-4{\int }_{0}^{\frac{1}{\sqrt{2}}}(\frac{-2x}{1-{x}^{2}})dx$$=−4∫02​1​​(1−x2−2x​)dx

$$=-4{[{\mathrm{log}}_{e}(1-{x}^{2})]}_{0}^{\frac{1}{\sqrt{2}}}$$=−4[loge​(1−x2)]02​1​​

$$=-4{\mathrm{log}}_{e}(1-\frac{1}{2})$$=−4loge​(1−21​)

$$=4{\mathrm{log}}_{e}2$$=4loge​2

$$={\mathrm{log}}_{e}16$$=loge​16