calculate-0-dt-3-t-2-1-t-dt-

Question Number 44173 by abdo.msup.com last updated on 22/Sep/18

c a l c u l a t e \int_{0}^{\infty} \frac{d t}{(3 + t^{2}) \sqrt{1 + t}} d t

Commented by maxmathsup by imad last updated on 23/Sep/18

l e t I = \int_{0}^{\infty} \frac{d t}{(3 + t^{2}) \sqrt{1 + t}} c h a n g e m e n t \sqrt{1 + t} = x g i v e 1 + t = x^{2} a n d

I = \int_{0}^{\infty} \frac{2 x d x}{(3 + {(x^{2} - 1)}^{2}) x} = \int_{0}^{\infty} \frac{2 d x}{3 + x^{4} - 2 x^{2} + 1} = \int_{0}^{\infty} \frac{2 d x}{x^{4} - 2 x^{2} + 4}

= \int_{0}^{\infty} \frac{2 d x}{(x^{4} + 4 - 2 x^{2})} = \int_{0}^{\infty} \frac{2 d x}{{(x^{2} + 2)}^{2} - 6 x^{2}} = \int_{0}^{\infty} \frac{2 d x}{(x^{2} + 2 + \sqrt{6} x) (x^{2} + 2 - \sqrt{6} x)}

= \int_{0}^{\infty} \frac{2 d x}{(x^{2} + x \sqrt{6} + 2) (x^{2} - \sqrt{6} x + 2)} l e t d e c o m p o s e

F (x) = \frac{2}{(x^{2} + x \sqrt{6} + 2) (x^{2} - \sqrt{6} x + 2)} \Rightarrow F (x) = \frac{a x + b}{x^{2} + x \sqrt{6} + 2} + \frac{c x + d}{x^{2} - x \sqrt{6} + 2}

F (- x) = F (x) \Rightarrow \frac{- a x + b}{x^{2} - x \sqrt{6} + 2} + \frac{- c x + d}{x^{2} + x \sqrt{6} + 2} = F (x) \Rightarrow c = - a a n d b = d

\Rightarrow F (x) = \frac{a x + b}{x^{2} + x \sqrt{6} + 2} + \frac{- a x + b}{x^{2} - x \sqrt{6} + 2}

F (0) = \frac{1}{2} = \frac{b}{2} + \frac{b}{2} = b \Rightarrow b = \frac{1}{2}

F (1) = \frac{a + \frac{1}{2}}{3 + \sqrt{6}} + \frac{- a + \frac{1}{2}}{3 - \sqrt{6}} = \frac{2}{3} \Rightarrow \frac{2 a + 1}{2 (3 + \sqrt{6})} + \frac{1}{2} \frac{- 2 a + 1}{3 - \sqrt{6}} = \frac{2}{3} \Rightarrow

\frac{2 a + 1}{3 + \sqrt{6}} + \frac{- 2 a + 1}{3 - \sqrt{6}} = \frac{4}{3} \Rightarrow \frac{2 a (3 - \sqrt{6}) + 3 - \sqrt{6} - 2 a (3 + \sqrt{6}) + 3 + \sqrt{6}}{3} = \frac{4}{3} \Rightarrow

- 4 a \sqrt{6} + 6 = 4 \Rightarrow - 4 a \sqrt{6} = - 2 \Rightarrow a = \frac{1}{2 \sqrt{6}} \Rightarrow

F (x) = \frac{\frac{1}{2 \sqrt{6}} x + \frac{1}{2}}{x^{2} + x \sqrt{6} + 2} + \frac{- \frac{1}{2 \sqrt{6}} x + \frac{1}{2}}{x^{2} - x \sqrt{6} + 2} = \frac{1}{2 \sqrt{6}} {\frac{x + \sqrt{6}}{x^{2} + x \sqrt{6} + 2} + \frac{- x + \sqrt{6}}{x^{2} - x \sqrt{6} + 2}} \Rightarrow

2 \sqrt{6} \int F (x) = \frac{1}{2} \int \frac{2 x + 2 \sqrt{6}}{x^{2} + x \sqrt{6} + 2} d x - \frac{1}{2} \int \frac{2 x - 2 \sqrt{6}}{x^{2} - x \sqrt{6} + 2} d x

= \frac{1}{2} l n ∣ \frac{x^{2} + x \sqrt{6} + 2}{x^{2} - x \sqrt{6} + 2} ∣ + \sqrt{6} \int \frac{d x}{x^{2} + x \sqrt{6} + 2} + \sqrt{6} \int \frac{d x}{x^{2} - x \sqrt{6} + 2} b u t

\int \frac{d x}{x^{2} + x \sqrt{6} + 2} = \int \frac{d x}{x^{2} + 2 \frac{\sqrt{6}}{2} x + \frac{3}{2} + 2 - \frac{3}{2}} = \int \frac{d x}{{(x + \frac{\sqrt{6}}{2})}^{2} + \frac{1}{2}}

=_{x + \frac{\sqrt{6}}{2} = \frac{u}{\sqrt{2}}} 2 \int \frac{1}{1 + u^{2}} \frac{d u}{\sqrt{2}} = \sqrt{2} a r c t a n (x \sqrt{2} + \sqrt{3}) + c_{1} a l s o

\int \frac{d x}{x^{2} - x \sqrt{6} + 2} = \sqrt{2} a r c t a n (x \sqrt{2} - \sqrt{3}) + c \Rightarrow

2 \sqrt{6} \int_{0}^{+ \infty} F (x) d x = {[\frac{1}{2} l n ∣ \frac{x^{2} + x \sqrt{6} + 2}{x^{2} - x \sqrt{6} + 2} ∣]}_{0}^{+ \infty} + \sqrt{6} {[\sqrt{2} a r c t a n (x \sqrt{2} + \sqrt{3})]}_{0}^{+ \infty}

+ \sqrt{6} {[\sqrt{2} a r c t a n (x \sqrt{2} - \sqrt{3})]}_{0}^{+ \infty}

= \sqrt{6} {\frac{π}{2} \sqrt{2} - \sqrt{2} a r c t a n (\sqrt{3})} + \sqrt{6} {\frac{π}{2} \sqrt{2} + \sqrt{2} a r c t a n (\sqrt{3})}

= π \sqrt{3} + π \sqrt{3} = 2 π \sqrt{3} \Rightarrow \int_{0}^{\infty} F (x) d x = \frac{2 π \sqrt{3}}{2 \sqrt{6}} = \frac{π}{\sqrt{2}} \Rightarrow I = \frac{π}{\sqrt{2}} .

Answered by tanmay.chaudhury50@gmail.com last updated on 23/Sep/18

y^{2} = 1 + t d t = 2 y d y

\int_{1}^{\infty} \frac{2 y d y}{y (3 + y^{4} - 2 y^{2} + 1)}

\int_{1}^{\infty} \frac{2 d y}{(y^{4} - 2 y^{2} + 4)}

\int_{1}^{\infty} \frac{\frac{2}{y^{2}}}{(y^{2} - 2 + \frac{4}{y^{2}})} d y

d (y + \frac{2}{y}) = 1 - \frac{2}{y^{2}}

d (y - \frac{2}{y}) = 1 + \frac{2}{y^{2}}

\frac{1}{2} \int_{1}^{\infty} \frac{(1 + \frac{2}{y^{2}}) - (1 - \frac{2}{y^{2}})}{(y^{2} + \frac{4}{y^{2}} - 2)} d y

\frac{1}{2} \int_{1}^{\infty} \frac{d (y - \frac{2}{y}) - d (y + \frac{2}{y})}{(y^{2} + \frac{4}{y^{2}} - 2)}

\frac{1}{2} [\int_{1}^{\infty} \frac{d (y - \frac{2}{y})}{{(y - \frac{2}{y})}^{2} + 2 . y . \frac{2}{y} - 2} - \int_{1}^{\infty} \frac{d (y + \frac{2}{y})}{{(y + \frac{2}{y})}^{2} - 2 . y . \frac{2}{y} + 2}]

\frac{1}{2} [\int_{- 1}^{\infty} \frac{{d x}_{1}}{x_{1}^{2} + 2} + \int_{3}^{\infty} \frac{{d x}_{2}}{x_{2}^{2} - 2}] n o w u s e f o r m u l a

\frac{1}{2} [\frac{1}{\sqrt{2}} ∣ {t a n}^{- 1} (\frac{x_{1}}{\sqrt{2}}) ∣_{- 1}^{\infty} + \frac{1}{2 \sqrt{2}} ∣ l n (\frac{x_{2} - \sqrt{2}}{x_{2} + \sqrt{2}}) ∣_{3}^{\infty}]

\frac{1}{2} [\frac{1}{\sqrt{2}} (\frac{Π}{2} - {{t a n}^{- 1} (\frac{- 1}{\sqrt{2}})} + \frac{1}{2 \sqrt{2}} ∣ l n (\frac{1 - \frac{\sqrt{2}}{x_{2}}}{1 + \frac{\sqrt{2}}{x_{2}}}) ∣_{3}^{\infty}]

\frac{1}{2} [\frac{1}{\sqrt{2}} {(\frac{Π}{2} + {t a n}^{- 1} (\frac{1}{\sqrt{2}})} + \frac{1}{2 \sqrt{2}} (0 - l n ∣ (\frac{3 - \sqrt{2}}{3 + \sqrt{2}} ∣)]

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