Question

*pls intergrate this and Don't spam*
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Answer 1

$\begin{gathered}\Large{\bold{\pink{\underline{Formula \: Used \::}}}} \end{gathered}$

$1. \: \boxed{ \red{ \bf \:\dfrac{d}{dx} x \: = 1}}$

$2. \: \boxed{ \red{ \bf \:\dfrac{d}{dx} \dfrac{1}{x} \: = - \dfrac{1}{ {x}^{2} } }}$

$3. \: \boxed{ \red{ \bf \: \int\dfrac{dx}{ {x}^{2} + {a}^{2} } = \dfrac{1}{a} {tan}^{ - 1}\dfrac{x}{a} + c }}$

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$\large\underline\purple{\bold{Solution :- }}$

$:  \implies \tt \:Let \: I \: = \int \: \dfrac{ {x}^{2} + 1 }{ {x}^{4} + 1 } dx$

$\pink{ \sf \: on \: dividing \: numerator \: and \: denominator \: by \: {x}^{2} }$

$:  \implies \tt \: I = \int \: \dfrac{1 + \dfrac{1}{ {x}^{2} } }{ {x}^{2} + \dfrac{1}{ {x}^{2} } } dx$

$:  \implies \tt \: I = \int \: \dfrac{1 + \dfrac{1}{ {x}^{2} } }{ {x}^{2} + \dfrac{1}{ {x}^{2} } - 2 + 2 } dx$

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

$\red{ \sf \: Put \: x - \dfrac{1}{x} = y}$

$:  \implies \tt \: \dfrac{dy}{dx} = 1 + \dfrac{1}{ {x}^{2} }$

$:  \implies \tt \: dy \: = (1 + \dfrac{1}{ {x}^{2} } )dx$

$:  \implies \tt \: I = \int \: \dfrac{dy}{ {y}^{2} + 2 }$

$:  \implies \tt \: I = \int \: \dfrac{dy}{ {y}^{2} + {( \sqrt{2}) }^{2} }$

$:  \implies \tt \: I = \dfrac{1}{ \sqrt{2} } {tan}^{ - 1} \bigg(\dfrac{y}{ \sqrt{2} } \bigg) + c$

$:  \implies \tt \: I = \dfrac{1}{ \sqrt{2} } {tan}^{ - 1} \bigg(\dfrac{x - \dfrac{1}{x} }{ \sqrt{2} } \bigg) + c$

$:  \implies \tt \: I = \dfrac{1}{ \sqrt{2} } {tan}^{ - 1} \bigg(\dfrac{ {x}^{2} - 1 }{ \sqrt{2} x} \bigg) + c$