Question

Integration of 1/1+x^4​

Answer 1

Answer:

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Answer 2

$\int\limits \: \frac{1}{1 + {x}^{4} } dx \\ = \frac{1}{2} \int\limits \frac{( { {x}^{2} + 1) - ( {x}^{2} - 1)}^{2} }{1 + {x}^{4} } dx \\ \frac{1}{2} \int\limits \: dx - \frac{1}{2} \int\limits \frac{ {x}^{2} - 1}{1 + {x}^{4} } dx$

$\frac{x}{2} - \frac{1}{2} \int\limits \frac{1 - \frac{1}{ {x}^{2} } }{ \frac{1}{ {x}^{2} + {x}^{2} } } dx$

$let \: z = x + \frac{1}{x} \\ then \: using \: diffrentiation \\ dz = (1 - \frac{1}{ {x}^{2} } )dx$

$\frac{x}{2} - \frac{1}{2} \int\limits \frac{dz}{ {z}^{2} - 2} dz \\ \frac{x}{2} - \frac{1}{4 \sqrt{2} } log( \frac{z - \sqrt{2} }{z + \sqrt{2} } ) + c$

$\frac{x}{2} - \frac{1}{4 \sqrt{2} } log( \frac{ {x}^{2} - \sqrt{2}x + 1 }{ {x}^{2} + \sqrt{2}x + 1 } ) + c$

$\frac{x}{2} - \frac{1}{2} \int\limits \frac{1 - \frac{1}{ {x}^{2} } }{ {(1 + x)}^{2} \: - 2 }dx$