Question

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

Answer 1

Step 1. Integrate the given equation:

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

$={\int }_{\frac{1}{-\sqrt{2}}}^{\frac{1}{\sqrt{2}}}{\left(\frac{1-x}{1+x}-\frac{1+x}{1-x}\right)}^{2}dx$ $\left[\mathbf{\because }{\mathit{a}}^{\mathbf{2}}\mathbf{+}{\mathit{b}}^{\mathbf{2}}\mathbf{-}\mathbf{2}\mathit{a}\mathit{b}\mathbf{}\mathbf{=}\mathbf{}{\mathbf{(}\mathbf{a}\mathbf{-}\mathbf{b}\mathbf{)}}^{\mathbf{2}}\right]$

$=2{\int }_0^{\frac{1}{\sqrt{2}}}{\left(\frac{1-x}{1+x}-\frac{1+x}{1-x}\right)}^{2}dx$

$=2{\int }_0^{\frac{1}{\sqrt{2}}}\left(\frac{1+x^2-2x-1-x^2-2x}{1-x^2}\right)dx$

$=2{\int }_0^{\frac{1}{\sqrt{2}}}\left(\frac{-4x}{1-x^2}\right)dx$ ……(1)

Let $1-x^2=t$

Step 2. Differentiate it with respect to $x$

$\Rightarrow$$-2xdx=dt$

$\Rightarrow$ $dx=\frac{dt}{-2x}$

Step 3. Put the value of $dx$ and $1-x^2$ in equation (1), we get

$\Rightarrow$$4{\int }_1^{\frac{1}{2}}\frac{dt}{t}$

$\Rightarrow$$4{\left[\log t\right]}_1^{\frac{1}{2}}$ $\left[\mathbf{\because }{\mathbf{\int }}_{}\frac{\mathbf{1}}{\mathbf{x}}\mathit{d}\mathit{x}\mathbf{}\mathbf{=}\mathbf{}\mathit{l}\mathit{o}\mathit{g}\mathbf{}\mathit{x}\right]$

$\Rightarrow 4\left[\log \frac{1}{2}-\log 1\right]$

$\Rightarrow \mathbf{4}\mathbf{}\mathit{l}\mathit{o}\mathit{g}\frac{\mathbf{1}}{\mathbf{2}}$ $\left[\mathbf{\because }\mathit{l}\mathit{o}\mathit{g}\mathbf{}\mathit{A}\mathbf{-}\mathit{l}\mathit{o}\mathit{g}\mathbf{}\mathit{B}\mathbf{}\mathbf{=}\mathbf{}\mathit{l}\mathit{o}\mathit{g}\mathbf{}\frac{\mathbf{A}}{\mathbf{B}}\right]$

Hence, the integration of given equation is $\mathbf{4}\mathbf{}\mathit{l}\mathit{o}\mathit{g}\frac{\mathbf{1}}{\mathbf{2}}$