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Question Number 83085 by mathmax by abdo last updated on 27/Feb/20

1) f i n d \int \frac{d x}{{(x^{2} + 1)}^{4}}

2) c a l c u l a t e \int_{0}^{\infty} \frac{d x}{{(x^{2} + 1)}^{4}}

Commented by abdomathmax last updated on 28/Feb/20

2) l e t I = \int_{0}^{\infty} \frac{d x}{{(x^{2} + 1)}^{4}} d x \Rightarrow 2 I = \int_{- \infty}^{+ \infty} \frac{d x}{{(x^{2} + 1)}^{4}}

l e t W (z) = \frac{1}{{(z^{2} + 1)}^{4}} \Rightarrow W (z) = \frac{1}{{(z - i)}^{4} {(z + i)}^{4}}

\int_{- \infty}^{+ \infty} W (z) d z = 2 i π R e s (W, i)

R e s (W, i) = {l i m}_{z \to i} \frac{1}{(4 - 1)!} {{(z - i)}^{4} W (z)}^{(3)}

= {l i m}_{z \to i} \frac{1}{6} {{(z + i)}^{- 4}}^{(3)}

= {l i m}_{z \to i} \frac{- 2}{3} {{(z + i)}^{- 5}}^{(2)}

= {l i m}_{z \to i} \frac{10}{3} {{(z + i)}^{- 6}}^{) 1)}

= {l i m}_{z \to i} (- 20) {(z + i)}^{- 7}

= \frac{- 20}{{(2 i)}^{7}} = \frac{- 20}{2^{7} \times i^{8 - 1}} = \frac{- 20 i}{2^{7}} = \frac{- 20 i}{2^{5} \times 4} = \frac{- 20 i}{32 \times 4}

= - \frac{4 \times 5 i}{32 \times 4} = - \frac{5 i}{32} \Rightarrow \int_{- \infty}^{+ \infty} W (z) d z = 2 i π \times (\frac{- 5 i)}{32})

= \frac{5 π}{16} = 2 I \Rightarrow I = \frac{5 π}{32}

Commented by mathmax by abdo last updated on 28/Feb/20

1) c o m p l e x m e t h o d l e t A = \int \frac{d x}{{(x^{2} + 1)}^{4}} \Rightarrow A = \int \frac{d x}{{(x - i)}^{4} {(x + i)}^{4}}

= \int \frac{d x}{{(\frac{x - i}{x + i})}^{4} {(x + i)}^{8}} l e t u s e t h e c h a n g e m e n t \frac{x - i}{x + i} = t \Rightarrow

x - i = x t + i t \Rightarrow (1 - t) x = i (1 + t) \Rightarrow x = i \frac{1 + t}{1 - t} \Rightarrow d x = i \frac{1 - t + (1 + t)}{{(1 - t)}^{2}} d t

= \frac{2 i}{{(t - 1)}^{2}} d t a l s o x + i = \frac{i + i t}{1 - t} + i = \frac{i + i t + i - i t}{1 - t} = \frac{2 i}{1 - t} \Rightarrow

A = \int \frac{2 i d t}{{(t - 1)}^{2} \times t^{4} \times {(\frac{2 i}{t - 1})}^{8}} = \frac{1}{{(2 i)}^{7}} \int \frac{{(t - 1)}^{8} d t}{{(t - 1)}^{2} t^{4}}

= \frac{i}{2^{7}} \int \frac{{(t - 1)}^{6}}{t^{4}} d t = \frac{i}{2^{7}} \int \frac{\sum_{k = 0}^{6} C_{6}^{k} {(- 1)}^{k} t^{k}}{t^{4}} d t

= \frac{i}{2^{7}} \int \sum_{k = 0}^{6} {(- 1)}^{k} C_{6}^{k} t^{k - 4} d t

= \frac{i}{2^{7}} {\sum_{k = 0 a n d k \neq 3}^{6} {(- 1)}^{k} C_{6}^{k} \times \frac{1}{k - 3} t^{k - 3} - C_{6}^{3} l n (t)} + C

\Rightarrow A = \frac{i}{2^{7}} {\sum_{k = 0 a n d k \neq 3}^{6} \frac{{(- 1)}^{k} C_{6}^{k}}{k - 3} {(\frac{x - i}{x + i})}^{k - 3} - C_{6}^{3} l n (\frac{x - i}{x + i})} + C

Answered by mind is power last updated on 27/Feb/20

\int \frac{d x}{(x^{2} + a)} = f (a)

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

\int \frac{d x}{x^{2} + a} = \frac{a r c t a n (\frac{x}{\sqrt{a}})}{\sqrt{a}} + c = f (a)

\int \frac{d x}{{(x^{2} + a)}^{4}} = - \frac{f^{4} (a)}{6}