dx-1-x-4-1-2-dx-1-ix-2-1-2-dx-1-ix-2-1-2-e-i-pi-4-d-e-i-pi-4-x-1-e-i-pi-4-x-2-1-2-e-i-pi-4-d-e-i-pi-4-x-1-e-i-pi-4-x-2-1-2-e-i-pi-4-ta

Question Number 132799 by Ñï= last updated on 16/Feb/21

\int \frac{d x}{1 + x^{4}}

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

= \frac{1}{2} e^{- i \frac{π}{4}} \int \frac{d (e^{i \frac{π}{4}} x)}{1 + {(e^{i \frac{π}{4}} x)}^{2}} + \frac{1}{2} e^{i \frac{π}{4}} \int \frac{d (e^{- i \frac{π}{4}}) x}{1 + {(e^{- i \frac{π}{4}} x)}^{2}}

= \frac{1}{2} e^{- i \frac{π}{4}} {t a n}^{- 1} (e^{i \frac{π}{4}} x) + \frac{1}{2} e^{i \frac{π}{4}} {t a n}^{- 1} (e^{- i \frac{π}{4}} x) + C

= Q u e s t i o n i s h o w t o c o n t i n u e f i n i s h i t ?

Commented by Ar Brandon last updated on 16/Feb/21

x^{4} + 1 = {(x^{2} + 1)}^{2} - {2 x}^{2} = {(x^{2} + 1)}^{2} - {(\sqrt{2} x)}^{2}

= (x^{2} - \sqrt{2} x + 1) (x^{2} + \sqrt{2} x + 1)

\frac{1}{x^{4} + 1} = \frac{ax + b}{x^{2} - \sqrt{2} x + 1} + \frac{cx + d}{x^{2} + \sqrt{2} x + 1}

a + c = 0, \sqrt{2} a + b - \sqrt{2} c + d = 0, a + \sqrt{2} b + c - \sqrt{2} d = 0, b + d = 1

b = \frac{1}{2} = d, a = - \frac{1}{2 \sqrt{2}}, c = \frac{1}{2 \sqrt{2}}

Commented by MJS_new last updated on 16/Feb/21

\arctan (a + b i) =

= \frac{1}{2} (π sign a + \arctan \frac{b - 1}{a} - \arctan \frac{b + 1}{a}) + \frac{i}{4} \ln \frac{a^{2} + {(b + 1)}^{2}}{a^{2} + {(b - 1)}^{2}}

{that}^{'} s why I^{'} d rather not \dots

Answered by MJS_new last updated on 16/Feb/21

\int \frac{d x}{x^{4} + 1} = \frac{1}{4} \int \frac{\sqrt{2} x + 2}{x^{2} + \sqrt{2} x + 1} - \frac{\sqrt{2} x - 2}{x^{2} - \sqrt{2} x + 1} d x =

= \int \frac{2 x + \sqrt{2}}{4 \sqrt{2} (x^{2} + \sqrt{2} x + 1)} + \frac{1}{4 (x^{2} + \sqrt{2} x + 1)} - \frac{2 x - \sqrt{2}}{4 \sqrt{2} (x^{2} + \sqrt{2} x + 1)} + \frac{1}{4 (x^{2} - \sqrt{2} x + 1)} d x =

= \frac{\sqrt{2}}{8} \ln (x^{2} + \sqrt{2} x + 1) + \frac{\sqrt{2}}{4} \arctan (\sqrt{2} x + 1) - \frac{\sqrt{2}}{8} \ln (x^{2} - \sqrt{2} x + 1) + \frac{\sqrt{2}}{4} \arctan (\sqrt{2} x - 1) + C

Commented by Ñï= last updated on 16/Feb/21

I w a n t t o k n o w h o w t o u s e arc t a n f u c t i o n t o g e t i t b a c k .

T h i s m e t h o d I a l r e a d y g e t i t . s i r .

Answered by Ar Brandon last updated on 16/Feb/21

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

= \frac{1}{2} \int \frac{1 + \frac{1}{x^{2}}}{x^{2} + \frac{1}{x^{2}}} dx - \frac{1}{2} \int \frac{1 - \frac{1}{x^{2}}}{x^{2} + \frac{1}{x^{2}}} dx = \frac{1}{2} \int \frac{1 + \frac{1}{x^{2}}}{{(x - \frac{1}{x})}^{2} + 2} dx - \frac{1}{2} \int \frac{1 - \frac{1}{x^{2}}}{{(x + \frac{1}{x})}^{2} - 2} dx

= \frac{1}{2} \int \frac{du}{u^{2} + 2} - \frac{1}{2} \int \frac{dv}{v^{2} - 2} = \frac{\tan^{- 1} (u / \sqrt{2})}{2 \sqrt{2}} + \frac{\tanh^{- 1} (v / \sqrt{2})}{2 \sqrt{2}} + C

= \frac{1}{2 \sqrt{2}} \tan^{- 1} (\frac{x^{2} - 1}{\sqrt{2} x}) + \frac{1}{4 \sqrt{2}} \ln ∣ \frac{x^{2} + x + 1}{x^{2} - x + 1} ∣ + C

Answered by mathmax by abdo last updated on 16/Feb/21

for that find \arctan (x + iy) \dots