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Answer 1

★ S O L U T I O N :

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$\sf :\implies \: (1) = \: \int \: \dfrac{1}{ {x}^{4} - 1} dx$

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As we know that,

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Formula of (a² - b²) = (a + b) (a - b)

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Using this formula, we get,

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$\sf :\implies \: \int \dfrac{1}{( {x}^{2} + 1)( {x}^{2} - 1)} dx$

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$\sf :\implies \: \dfrac{1}{2} \int \: \dfrac{( {x}^{2} + 1) - ( {x}^{2} - 1) }{( {x}^{2} - 1)( {x}^{2} + 1) } dx$

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$\sf :\implies \: \dfrac{1}{2} \int \: \bigg(\dfrac{1}{( {x}^{2} - 1)} \: \: - \: \: \dfrac{1}{( {x}^{2} + 1)} \bigg)dx$

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$\sf :\implies \: \dfrac{1}{2} \int \: \dfrac{1}{( {x}^{2} - 1)} dx \: \: - \: \: \dfrac{1}{2} \int \: \dfrac{1}{( {x}^{2} + 1)} dx$

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By applying the formula,

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$\sf :\implies \: \int \: \dfrac{1}{ {x}^{2} - {a}^{2} } dx = \dfrac{1}{2a} log \bigg( \dfrac{x - a}{x + a} \bigg ) + c$

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$\sf :\implies \: \dfrac{1}{ {x}^{2} + 1} = \tan {}^{ - 1} (x)$

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$\sf :\implies \: \int \: \dfrac{1}{4} log \bigg |\dfrac{x - 1}{x + 1} \bigg | \: - \dfrac{1}{2} \: \tan {}^{ - 1} (x) \: + c$

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$\sf :\implies \:(2) = \int \: \dfrac{1}{ ({x}^{2} - {a}^{2}) } dx$

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As we know that,

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

Formula of (a² - b²) = (a + b) (a - b)

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Apply this formula, we get,

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$\sf :\implies \: \int \: \dfrac{1}{(x - a)(x + a)}dx$

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$\sf :\implies \: \dfrac{1}{2a} \int \: \dfrac{(x + a) - (x - a)}{(x + a)(x - a)} dx$

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$\sf :\implies \: \dfrac{1}{2a} \int \dfrac{(x + a)}{(x + a)(x - a)} dx \: \: - \: \: \dfrac{1}{2a} \int \: \dfrac{(x - a)}{(x + a)(x - a) } dx$

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$\sf :\implies \: \int \: \dfrac{1}{2a} \bigg( \dfrac{dx}{(x - a)} \: \: - \: \: \dfrac{dx}{(x + a)} \bigg)$

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$\sf :\implies \: \dfrac{1}{2a} \bigg( log(x - a) - log(x + a) \bigg) + c$

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$\sf :\implies \: \dfrac{1}{2a} log \bigg | \dfrac{x - a}{x + a} \bigg | + c$

Answer 2

Given Question

$\bf \: Evaluate : \: \displaystyle \int \sf \: \dfrac{dx}{ {x}^{4} - 1 }$

Answer

Identities Used :-

$\boxed{ \red{\displaystyle \int \sf \: \dfrac{dx}{ {x}^{2} + {a}^{2}} = \dfrac{1}{a} {tan}^{ - 1}\dfrac{x}{a} + c}}$

$\boxed{ \red{\displaystyle \int \sf \: \dfrac{dx}{ {x}^{2} - {a}^{2}} = \dfrac{1}{2a}log\dfrac{x - a}{x + a} + c}}$

$\boxed{ \blue{ \sf \: {x}^{2} - {y}^{2} = (x + y)(x - y)}}$

Step by step explanation :-

$\rm :\longmapsto\:\displaystyle \int \sf \: \dfrac{dx}{ {x}^{4} - 1}$

$\sf \: \: = \: \displaystyle \int \sf \: \dfrac{dx}{ {x}^{4} - 1}$

$\sf \: \: = \: \displaystyle \int \sf \: \dfrac{dx}{ ({x}^{2})^{2} - 1}$

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

On multiply and divide by 2, we get

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

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

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

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

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

$\sf \: \: = \:\dfrac{1}{2} \displaystyle \int \sf \: \dfrac{dx}{ {x}^{2} - 1 } - \dfrac{1}{2}\displaystyle \int \sf \: \dfrac{dx}{ {x}^{2} + 1}$

$\sf \: \: = \:\dfrac{1}{2} \times \dfrac{1}{2 \times 1}log |\dfrac{x - 1}{x + 1} | - \dfrac{1}{2} \times \dfrac{1}{1} {tan}^{ - 1}\dfrac{x}{1} + c$

$\sf \: \: = \:\dfrac{1}{4}log |\dfrac{x - 1}{x + 1} | - \dfrac{1}{2} {tan}^{ - 1}x + c$

Additional Information :-

$\boxed{ \red{\displaystyle \int \sf \: \dfrac{dx}{ {a}^{2} - {x}^{2}} = \dfrac{1}{2a}log\dfrac{a + x}{a - x} + c}}$

$\boxed{ \red {\displaystyle \int \sf \: \dfrac{dx}{ \sqrt{ {a}^{2} - {x}^{2}}} = {sin}^{ - 1}\dfrac{x}{a} + c }}$

$\boxed{ \red {\displaystyle \int \sf \: \dfrac{dx}{ \sqrt{ {x}^{2} - {a}^{2}}} = log |x + \sqrt{ {x}^{2} - {a}^{2} } | + c}}$

$\boxed{ \red {\displaystyle \int \sf \: \dfrac{dx}{ \sqrt{ {x}^{2} + {a}^{2}}} = log |x + \sqrt{ {x}^{2} + {a}^{2} } | + c}}$