Question

integration of √[(x+2 / x-2 )² + (x-2 / x+2)² - 2]​

Answer 1

$\large\underline{\sf{Solution-}}$

The given integral is

$\rm :\longmapsto\:\displaystyle\int_{-1}^1\sf \sqrt{ {\bigg(\dfrac{x + 2}{x - 2} \bigg) }^{2} + {\bigg(\dfrac{x - 2}{x + 2} \bigg) }^{2} - 2 } \: dx$

We know,

$\boxed{ \sf{ \:{\bigg(x - \dfrac{1}{x} \bigg) }^{2} = {x}^{2} + \frac{1}{ {x}^{2} } - 2}}$

So, using this identity we get

$\rm \: = \: \:\displaystyle\int_{-1}^1\sf \sqrt{ {\bigg(\dfrac{x + 2}{x - 2} - \dfrac{x - 2}{x + 2} \bigg) }^{2} } \: dx$

$\rm \: = \: \:\displaystyle\int_{-1}^1\sf \sqrt{ {\bigg(\dfrac{ {(x + 2)}^{2} - {(x - 2)}^{2} }{(x - 2)(x + 2)} \bigg) }^{2} } \: dx$

$\rm \: = \: \:\displaystyle\int_{-1}^1\sf {\bigg(\dfrac{ {(x + 2)}^{2} - {(x - 2)}^{2} }{(x - 2)(x + 2)} \bigg) } \: dx$

$\rm \: = \: \:\displaystyle\int_{-1}^1\sf {\bigg(\dfrac{ {x}^{2} + 4 + 4x - {x}^{2} - 4 + 4x}{ {x}^{2} - 4} \bigg) } \: dx$

$\rm \: = \: \:\displaystyle\int_{-1}^1\sf {\bigg(\dfrac{8x}{ {x}^{2} - 4} \bigg) } \: dx$

$\rm \: = \: 4\:\displaystyle\int_{-1}^1\sf {\bigg(\dfrac{2x}{ {x}^{2} - 4} \bigg) } \: dx$

We know,

$\boxed{ \sf{ \:\displaystyle\int\sf \frac{f'(x)}{f(x)} \: dx = log |f(x)| + c}}$

Here,

$\boxed{ \sf{ \: \frac{d}{dx}( {x}^{2} - 4) = 2x}}$

So, using these, we get

$\rm \:  =  \:  \:4 \: log | {x}^{2} - 4|_{-1}^1$

$\rm \:  =  \:  \:4 \: log | {1}^{2} - 4| - 4 \: log | {( - 1)}^{2} - 4|$

$\rm \:  =  \:  \:4 \: log |1 - 4| - 4 \: log |1 - 4|$

$\rm \:  =  \:  \:4 \: log |3| - 4 \: log |3|$

$\rm \:  =  \:  \:0$

Additional Information :-

$\boxed{ \sf{ \:\displaystyle\int_{a}^b\sf f(x) \: dx = - \displaystyle\int_{b}^a\sf f(x) \: dx}}$

$\boxed{ \sf{ \:\:\displaystyle\int_{a}^b\sf f(x) \: dx = \:\displaystyle\int_{a}^b\sf f(y) \: dy }}$

$\boxed{ \sf{ \:\:\displaystyle\int_{a}^b\sf f(x) \: dx = \:\displaystyle\int_{a}^b\sf f(a + b - x) \: dy }}$

$\boxed{ \sf{ \:\:\displaystyle\int_{0}^a\sf f(x) \: dx = \:\displaystyle\int_{0}^a\sf f(a - x) \: dy }}$