let-f-x-arctan-1-ix-determine-Re-f-x-and-Im-f-x-dx-

Question Number 59182 by maxmathsup by imad last updated on 05/May/19

l e t f (x) = a r c t a n (1 + i x)

d e t e r m i n e R e (f (x)) a n d I m (f (x)) d x

Commented by maxmathsup by imad last updated on 10/Jun/19

w e h a v e f^{^{'}} (x) = \frac{i}{1 + {(1 + i x)}^{2}} = \frac{i}{1 + 1 + 2 i x - x^{2}} = \frac{i}{- x^{2} + 2 i x + 2}

= \frac{- i}{x^{2} - 2 i x - 2} = \frac{- i}{x^{2} - 2 - 2 i x} = \frac{- i (x^{2} - 2 + 2 i x)}{{(x^{2} - 2)}^{2} + 4 x^{2}} = \frac{2 x}{{(x^{2} - 2)}^{2} + 4 x^{2}} - i \frac{x^{2}}{{(x^{2} - 2)}^{2} + 4 x^{2}}

\Rightarrow f (x) = \int \frac{2 x d x}{{(x^{2} - 2)}^{2} + 4 x^{2}} + c_{0} - i (\int \frac{x^{2} d x}{{(x^{2} - 2)}^{2} + 4 x^{2}} + c_{1}) \Rightarrow

R e (f (x)) = \int \frac{2 x d x}{{(x^{2} - 2)}^{2} + 4 x^{2}} + c_{0} a n d I m (f (x)) = \int \frac{x^{2} d x}{{(x^{2} - 2)}^{2} + 4 x^{2}} + c_{1}

c_{0} a n d c_{1} r e a l s

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

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

F (x) = \frac{2 x}{x^{4} + 4} = \frac{2 x}{(x^{2} - 2 x + 2) (x^{2} + 2 x + 2)} = \frac{a x + b}{x^{2} - 2 x + 2} + \frac{c x + d}{x^{2} + 2 x + 2} \Rightarrow

\int F (x) d x [= \int \frac{a x + b}{x^{2} - 2 x + 2} d x + \int \frac{c x + d}{x^{2} + 2 x + 2} d x = \dots . b e c o n t i n u e d \dots .

Commented by mathmax by abdo last updated on 22/Jun/19

\int \frac{2 x d x}{{(x^{2} - 2)}^{2} + 4 x^{2}} =_{x^{2} = t} \int \frac{d t}{{(t - 2)}^{2} + 4 t} = \int \frac{d t}{t^{2} - 4 t + 4 + 4 t} = \int \frac{d t}{4 + t^{2}}

=_{t = 2 u} \int \frac{2 d u}{4 + 4 u^{2}} = \frac{1}{2} \int \frac{d u}{1 + u^{2}} = \frac{1}{2} a r c t a n (u) + c = \frac{1}{2} a r c t a n (\frac{x^{2}}{2}) + c \Rightarrow

R e (f (x)) = \frac{1}{2} a r c t a n (\frac{x^{2}}{2}) + c_{0}