find-the-integral-dx-x-4-a-4-by-complex-number-

Question Number 209521 by mokys last updated on 12/Jul/24

f i n d t h e i n t e g r a l \int \frac{d x}{x^{4} + a^{4}} b y c o m p l e x n u m b e r ?

Answered by mathmax last updated on 14/Jul/24

I = \int \frac{d x}{x^{4} - i^{2} a^{4}} = \int \frac{d x}{(x^{2} - {i a}^{2}) (x^{2} + {i a}^{2})}

= \frac{1}{2 {i a}^{2}} \int (\frac{1}{x^{2} - {i a}^{2}} - \frac{1}{x^{2} + {i a}^{2}}) d x

= \frac{1}{2 {i a}^{2}} {\int \frac{d x}{x^{2} - {i a}^{2}} - \int \frac{d x}{x^{2} + {i a}^{2}}} d x

I_{1} = \int \frac{d x}{(x - \sqrt{i} a) (x + \sqrt{i} a)}

= \frac{1}{2 \sqrt{i} a} \int (\frac{1}{x - \sqrt{i} a} - \frac{1}{x + \sqrt{i} a)}) d x

= \frac{1}{2 \sqrt{i} a} l n (\frac{x - \sqrt{i} a}{x + \sqrt{i} a}) + c_{1}

I_{2} = \int \frac{d x}{x^{2} + {i a}^{2}} = \int \frac{d x}{x^{2} - (- i) a^{2}}

= \frac{1}{2 \sqrt{- i} a} \int (\frac{1}{x - \sqrt{- i} a} - \frac{1}{x + \sqrt{- i} a}) d x

= \frac{1}{2 \sqrt{- i} a} l n (\frac{x - \sqrt{- i} a}{x + \sqrt{- i} a}) + c_{2}

I = \frac{1}{2 {i a}^{2}} \times \frac{1}{2 \sqrt{i} a} l n (\frac{x - \sqrt{i} a}{x + \sqrt{i} a})

- \frac{1}{2 {i a}^{2}} \times \frac{1}{2 \sqrt{- i} a} l n (\frac{x - \sqrt{- i} a}{x + \sqrt{- i} a}) + C

= \frac{1}{4 a^{3} i \sqrt{i}} l n (\frac{x - \sqrt{i} a}{x + \sqrt{i} a}) - \frac{1}{4 a^{3} i \sqrt{- i}} l n (\frac{x - \sqrt{- i} a}{x + \sqrt{- i} a}) + C

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