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Answer 1

EXPLANATION.

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

As we know that,

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

Using this formula, we get.

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

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

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

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

$\sf \implies \: by \: applying \: the \: formula \\ \\ \sf \implies \: \int \: \frac{1}{ {x}^{2} - {a}^{2} } dx. = \frac{1}{2a} log \bigg( \frac{x - a}{x + a} \bigg ) + c. \\ \\ \sf \implies \: \frac{1}{ {x}^{2} + 1} = \tan {}^{ - 1} (x)$

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

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

As we know that,

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

Apply this formula, we get.

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

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

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

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

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

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

Answer 2

☆ Given Question:-

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

$\huge \orange{AηsωeR} ✍$

☆Formula used :-

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

$\bf \:\int \dfrac{1}{ {x}^{2} + {a}^{2} } dx = \dfrac{1}{a} {tan}^{ - 1} \dfrac{x}{a} + c$

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$\large \red{\bf \: Solution :- } ✍$

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

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

☆ Multiply and divide by 2, we get

$\bf \:  ⟼\dfrac{1}{2} \int \dfrac{2}{ ({x}^{2} - 1)( {x}^{2} + 1) }$

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

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

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

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

$\bf \:  ⟼ \dfrac{1}{2} \int \dfrac{1}{ {x}^{2} - 1} dx - \dfrac{1}{2} \int \dfrac{1}{ {x}^{2} + 1}$

$\bf \:  ⟼ \dfrac{1}{2} \times \dfrac{1}{2 \times 1} log(\dfrac{x - 1}{x + 1} ) - \dfrac{1}{2} {tan}^{ - 1} x + c$

$\bf \:  ⟼ \dfrac{1}{4} log(\dfrac{x - 1}{x + 1} ) - \dfrac{1}{2} {tan}^{ - 1} x + c$

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