4x-1-8x-3-x-2-x-4-2x-2-2-dx-

Question Number 120102 by benjo_mathlover last updated on 29/Oct/20

Θ = \int \frac{4 x^{- 1} + 8 x^{- 3}}{x^{2} \sqrt{x^{4} + 2 x^{2} + 2}} d x

Answered by mathmax by abdo last updated on 29/Oct/20

I = \int \frac{{4 x}^{- 1} + {8 x}^{- 3}}{x^{2} \sqrt{x^{4} + {2 x}^{2} + 2}} dx \Rightarrow I = \int \frac{{4 x}^{2} + 8}{x^{5} \sqrt{x^{4} + {2 x}^{2} + 2}} dx \Rightarrow

I =_{x^{2} = t} \int \frac{4 t + 8}{t^{\frac{5}{2}} \sqrt{t^{2} + 2 t + 2}} \frac{dt}{2 \sqrt{t}} = \int \frac{2 t + 4}{t^{3} \sqrt{t^{2} + 2 t + 2}} dt

= \int \frac{2 t + 4}{t^{3} \sqrt{{(t + 1)}^{2} + 1}} dt =_{t + 1 = sh (t)} \int \frac{2 (sht - 1) + 4}{{(sht - 1)}^{3} ch (t)} ch (t) dt

= \int \frac{2 sht + 2}{{sh}^{3} t - {3 sh}^{2} t + 3 sht - 1} dt

= \int \frac{2 \frac{e^{t} - e^{- t}}{2} + 2}{{(\frac{e^{t} - e^{- t}}{2})}^{3} - 3 {(\frac{e^{t} - e^{- t}}{2})}^{2} + 3 \frac{e^{t} - e^{- t}}{2} - 1} dt

=_{e^{t} = u} \int \frac{u - u^{- 1} + 2}{\frac{1}{8} {(u - u^{- 1})}^{3} - \frac{3}{4} {(u - u^{- 1})}^{2} + \frac{3}{2} (u - u^{- 1}) - 1} \frac{du}{u}

= 8 \int \frac{u - u^{- 1} + 2}{u {{(u - u^{- 1})}^{3} - 6 {(u - u^{- 1})}^{2} + 12 (u - u^{- 1}) - 8}} du

= 8 \int \frac{u - u^{- 1} + 2}{u {u^{3} - {3 u}^{2} u^{- 1} + {3 uu}^{- 2} + u^{- 3} - 6 (u^{2} - 2 + u^{- 1}) + 12 u - {12 u}^{- 1} - 8}} du

= 8 \int \frac{u - u^{- 1} + 2}{u {u^{3} - 3 u + {3 u}^{- 1} + u^{- 3} - {6 u}^{2} + 12 - {6 u}^{- 1} + 12 u - {12 u}^{- 1} - 8}}

= 8 \int \frac{u - u^{- 1} + 2}{u {u^{3} - 3 u + {3 u}^{- 1} + u^{- 3} - {6 u}^{2} + 12 u - {18 u}^{- 1} - 8}} du

= 8 \int \frac{u - u^{- 1} + 2}{u^{4} - {3 u}^{2} + 3 + u^{- 2} - {6 u}^{3} + {12 u}^{2} - 18 + 8 u} du

= 8 \int \frac{u^{3} - u + {2 u}^{2}}{u^{6} + {9 u}^{4} + {3 u}^{2} + 1 - {6 u}^{5} - {18 u}^{2} + {8 u}^{3}} du

= 8 \int \frac{u^{3} + {2 u}^{2} - u}{u^{6} - {6 u}^{5} + {9 u}^{4} + {8 u}^{3} - {15 u}^{2} + 1} du rest decompositon of

F (u) = \frac{u^{3} + {2 u}^{2} - u}{u^{6} - {6 u}^{5} + {9 u}^{4} + {8 u}^{3} - {15 u}^{2} + 1} \dots . be continued \dots .

Answered by TANMAY PANACEA last updated on 29/Oct/20

\int \frac{4 x^{- 1} + 8 x^{- 3}}{x^{2} \sqrt{1 + {(x^{2} + 1)}^{2}}} d x

4 \int \frac{x^{2} + 2}{x^{5} \sqrt{1 + {(x^{2} + 1)}^{2}}} d x

x^{2} + 1 = t a n a \to 2 x d x = {s e c}^{2} a d a

d x = \frac{{s e c}^{2} a d a}{2 \sqrt{- 1 + t a n a}}

4 \int \frac{1 + t a n a}{{(- 1 + t a n a)}^{\frac{5}{2}} \times s e c a} \times \frac{{s e c}^{2} a d a}{2 \sqrt{- 1 + t a n a}}

2 \int \frac{1 + t a n a}{{(- 1 + t a n a)}^{3}} \times s e c a d a

2 \int \frac{(s i n a + c o s a)}{c o s a {(s i n a - c o s a)}^{3}} \times {c o s}^{3} a \times \frac{d a}{c o s a}

2 \int c o s a \times \frac{s i n a + c o s a}{{(s i n a - c o s a)}^{3}} d a

\frac{I}{2} = c o s a \int \frac{s i n a + c o s a}{{(s i n a - c o s a)}^{3}} - \int [\frac{d}{d a} (c o s a) \int \frac{s i n a + c o s a}{{(s i n a - c o s a)}^{3}}] d a

l o o k

d (s i n a - c o s a) = c o s a + s i n a

s o \int \frac{s i n a + c o s a}{{(s i n a - c o s a)}^{3}} d a = \int \frac{d (s i n a - c o s a)}{{(s i n a - c o s a)}^{3}}

= \frac{1}{- 2 {(s i n a - c o s a)}^{2}}

\frac{I}{2} = c o s a \times \frac{1}{- 2 {(s i n a - c o s a)}^{2}} - \int \frac{- s i n a}{1} \times \frac{1}{- 2 {(s i n a - c o s a)}^{2}} d a

I = c o s a \times \frac{1}{- {(s i n a - c o s a)}^{2}} - \int \frac{s i n a}{{(s i n a - c o s a)}^{2}}

I = \frac{c o s a}{- {(s i n a - c o s)}^{2}} - I_{★}

I_{★} = \frac{1}{2} \int \frac{s i n a + c o s a + s i n a - c o s a}{{(s i n a - c o s a)}^{2}} d a

\frac{1}{2} \int \frac{d (s i n a - c o s a)}{{(s i n a - c o s a)}^{2}} + \frac{1}{2} \int \frac{d a}{s i n a - c o s a}

\frac{1}{2} \times \frac{- 1}{(s i n a - c o s a)} + \frac{1}{2} \int \frac{\sqrt{2} d a}{s i n (a - \frac{π}{4})}

\frac{- 1}{2 (s i n a - c o s a)} + \frac{1}{\sqrt{2}} l n t a n (\frac{a}{2} - \frac{π}{8}) + c

n o t e t a n a = 1 + x^{2}

p l s c h k p l s

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