Question

# ${\int }_{\frac{1}{-\sqrt{2}}}^{\frac{1}{\sqrt{2}}}\left[{\left(\frac{1-x}{1+x}\right)}^{2}+{\left(\frac{1+x}{1-x}\right)}^{2}-2\right]dx$

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Solution

## Step 1. Integrate the given equation:${\int }_{\frac{1}{-\sqrt{2}}}^{\frac{1}{\sqrt{2}}}\left[{\left(\frac{1-x}{1+x}\right)}^{2}+{\left(\frac{1+x}{1-x}\right)}^{2}-2\right]dx$$={\int }_{\frac{1}{-\sqrt{2}}}^{\frac{1}{\sqrt{2}}}{\left(\frac{1-x}{1+x}-\frac{1+x}{1-x}\right)}^{2}dx$ $\left[\mathbf{\because }{\mathbit{a}}^{\mathbf{2}}\mathbf{+}{\mathbit{b}}^{\mathbf{2}}\mathbf{-}\mathbf{2}\mathbit{a}\mathbit{b}\mathbf{}\mathbf{=}\mathbf{}{\mathbf{\left(}\mathbf{a}\mathbf{-}\mathbf{b}\mathbf{\right)}}^{\mathbf{2}}\right]$$=2{\int }_{0}^{\frac{1}{\sqrt{2}}}{\left(\frac{1-x}{1+x}-\frac{1+x}{1-x}\right)}^{2}dx$$=2{\int }_{0}^{\frac{1}{\sqrt{2}}}\left(\frac{1+{x}^{2}-2x-1-{x}^{2}-2x}{1-{x}^{2}}\right)dx$$=2{\int }_{0}^{\frac{1}{\sqrt{2}}}\left(\frac{-4x}{1-{x}^{2}}\right)dx$ ……(1)Let $1-{x}^{2}=t$Step 2. Differentiate it with respect to $x$$⇒$$-2xdx=dt$$⇒$ $dx=\frac{dt}{-2x}$Step 3. Put the value of $dx$ and $1-{x}^{2}$ in equation (1), we get$⇒$$4{\int }_{1}^{\frac{1}{2}}\frac{dt}{t}$$⇒$$4{\left[\mathrm{log}t\right]}_{1}^{\frac{1}{2}}$ $\left[\mathbf{\because }{\mathbf{\int }}_{}\frac{\mathbf{1}}{\mathbf{x}}\mathbit{d}\mathbit{x}\mathbf{}\mathbf{=}\mathbf{}\mathbit{l}\mathbit{o}\mathbit{g}\mathbf{}\mathbit{x}\right]$$⇒4\left[\mathrm{log}\frac{1}{2}-\mathrm{log}1\right]$$⇒\mathbf{4}\mathbf{}\mathbit{l}\mathbit{o}\mathbit{g}\frac{\mathbf{1}}{\mathbf{2}}$ $\left[\mathbf{\because }\mathbit{l}\mathbit{o}\mathbit{g}\mathbf{}\mathbit{A}\mathbf{-}\mathbit{l}\mathbit{o}\mathbit{g}\mathbf{}\mathbit{B}\mathbf{}\mathbf{=}\mathbf{}\mathbit{l}\mathbit{o}\mathbit{g}\mathbf{}\frac{\mathbf{A}}{\mathbf{B}}\right]$Hence, the integration of given equation is $\mathbf{4}\mathbf{}\mathbit{l}\mathbit{o}\mathbit{g}\frac{\mathbf{1}}{\mathbf{2}}$

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