find f(a) =∫_a ^(+∞) (dx/((1+x^2 )(√(x^2 −a^2 )))) with a>0


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Question Number 50423 by Abdo msup. last updated on 16/Dec/18

$f i n d f (a) = \int_{a}^{+ \infty} \frac{d x}{(1 + x^{2}) \sqrt{x^{2} - a^{2}}} w i t h a > 0$
Commented byAbdo msup. last updated on 17/Dec/18

$c h a n g e m e n t x = a c h (t) g i v e t = a r g c h (\frac{x}{a})$ $f (a) = \int_{0}^{+ \infty} \frac{a s h (t) d t}{(1 + a^{2} {c h}^{2} (t)) a s h (t)}$ $= \int_{0}^{\infty} \frac{d t}{1 + a^{2} \frac{1 + c h (2 t)}{2}} = \int_{0}^{\infty} \frac{2 d t}{2 + a^{2} + a^{2} \frac{e^{2 t} + e^{- 2 t}}{2}}$ $= \int_{0}^{\infty} \frac{4 d t}{4 + 2 a^{2} + a^{2} e^{2 t} + a^{2} e^{- 2 t}}$ $=_{e^{2 t} = u} \int_{1}^{\infty} \frac{4}{4 + 2 a^{2} + a^{2} u + a^{2} u^{- 1}} \frac{d u}{2 u}$ $= 2 \int_{1}^{\infty} \frac{d u}{(4 + 2 a^{2}) u + a^{2} u^{2} + a^{2}}$ $= \int_{1}^{\infty} \frac{2 d u}{a^{2} u^{2} + (4 + 2 a^{2}) u + a^{2}}$ $Δ^{^{'}} = {(2 + a^{2})}^{2} - a^{4} = 4 + 4 a^{2} > 0 \Rightarrow$ $u_{1} = \frac{- 2 - a^{2} + 2 \sqrt{1 + a^{2}}}{a^{2}}$ $u_{2} = \frac{- 2 - a^{2} - 2 \sqrt{1 + a^{2}}}{a^{2}}$ $F (u) = \frac{2}{(u - u_{1}) (u - u_{2})} = \frac{2}{u_{1} - u_{2}} (\frac{1}{u - u_{1}} - \frac{1}{u - u_{2}})$ $= \frac{a^{2}}{2 \sqrt{1 + a^{2}}} (\frac{1}{u - u_{1}} - \frac{1}{u - u_{2}}) \Rightarrow$ $\int_{1}^{\infty} F (u) d u = \frac{a^{2}}{2 \sqrt{1 + a^{2}}} {[l n ∣ \frac{u - u_{1}}{u - u_{2}} ∣]}_{1}^{+ \infty}$ $= \frac{a^{2}}{2 \sqrt{1 + a^{2}}} l n ∣ \frac{1 - u_{2}}{1 - u_{1}} ∣$ $= \frac{a^{2}}{2 \sqrt{1 + a^{2}}} l n ∣ \frac{1 - \frac{- 2 - a^{2} - 2 \sqrt{1 + a^{2}}}{a^{2}}}{1 - \frac{- 2 - a^{2} + 2 \sqrt{1 + a^{2}}}{a^{2}}} ∣$ $= \frac{a^{2}}{2 \sqrt{1 + a^{2}}} l n ∣ \frac{2 a^{2} + 2 + 2 \sqrt{1 + a^{2}}}{2 a^{2} + 2 - 2 \sqrt{1 + a^{2}}} ∣$ $= \frac{a^{2}}{2 \sqrt{1 + a^{2}}} l n ∣ \frac{a^{2} + 1 + \sqrt{1 + a^{2}}}{a^{2} + 1 - \sqrt{1 + a^{2}}} ∣ = f (a)$
	Answered by tanmay.chaudhury50@gmail.com last updated on 17/Dec/18
	$∫(dx/((1+x^2 )(√(x^2 −a^2 )))) t=(1/x) x=(1/t) dx=((−1)/t^2 )dt ∫((−dt)/(t^2 (1+(1/t^2 ))(√((1/t^2 )−a^2 )) )) ∫((−tdt)/((t^2 +1)(√(1−a^2 t^2 )))) =((−1)/a)∫((tdt)/((t^2 +1)(√((1/a^2 )−t^2 )) )) k=t^2 dk=2tdt =((−1)/(2a))∫(dk/((k+1)(√((1/a^2 )−k)))) p^2 =(1/a^2 )−k 2pdp=−dk =((−1)/(2a))∫((−2pdp)/(((1/a^2 )−p^2 +1)p)) =(1/a)∫(dp/(((1/a^2 )+1)−p^2 )) formula[ ∫(dx/(a^2 −x^2 ))=(1/(2a))ln(((a+x)/(a−x)))] =(1/a)×(1/(2(√(((1/a^2 )+1))) ))×ln((((√((1/a^2 )+1)) +p)/((√((1/a^2 )+1)) −p))) =(1/(2(√(1+a^2 ))))ln((((√((1/a^2 )+1)) +(√((1/a^2 )−k)) )/((√((1/a^2 )+1)) −(√((1/a^2 )−k)) ))) =(1/(2(√(1+a^2 ))))ln((((√(1+a^2 )) +(√(1−a^2 k)) )/((√(1+a^2 )) −(√(1−a^2 k))))) =(1/(2(√(1+a^2 ))))ln((((√(1+a^2 )) +(√(1−a^2 t^2 )))/((√(1+a^2 )) −(√(1−a^2 t^2 )) ))) =(1/(2(√(1+a^2 ))))∣ln((((√(1+a^2 )) +(√(1−(a^2 /x^2 ))))/((√(1+a^2 )) −(√(1−(a^2 /x^2 ))))))∣_a ^∞ =(1/(2(√(1+a^2 ))))[{ln((((√(1+a^2 )) +(√(1−0)))/((√(1+a^2 )) −(√(1−0)))))}−ln((((√(1+a^2 )) +(√0))/((√(1+a^2 )) −(√0))))}] =(1/(√(1+a^2 )))ln((((√(1+a^2 )) +1)/((√(1+a^2 )) −1)))$
	$\int \frac{d x}{(1 + x^{2}) \sqrt{x^{2} - a^{2}}}$ $t = \frac{1}{x} x = \frac{1}{t} d x = \frac{- 1}{t^{2}} d t$ $\int \frac{- d t}{t^{2} (1 + \frac{1}{t^{2}}) \sqrt{\frac{1}{t^{2}} - a^{2}}}$ $\int \frac{- t d t}{(t^{2} + 1) \sqrt{1 - a^{2} t^{2}}}$ $= \frac{- 1}{a} \int \frac{t d t}{(t^{2} + 1) \sqrt{\frac{1}{a^{2}} - t^{2}}}$ $k = t^{2} d k = 2 t d t$ $= \frac{- 1}{2 a} \int \frac{d k}{(k + 1) \sqrt{\frac{1}{a^{2}} - k}}$ $p^{2} = \frac{1}{a^{2}} - k 2 p d p = - d k$ $= \frac{- 1}{2 a} \int \frac{- 2 p d p}{(\frac{1}{a^{2}} - p^{2} + 1) p}$ $= \frac{1}{a} \int \frac{d p}{(\frac{1}{a^{2}} + 1) - p^{2}} f o r m u l a [\int \frac{d x}{a^{2} - x^{2}} = \frac{1}{2 a} l n (\frac{a + x}{a - x})]$ $= \frac{1}{a} \times \frac{1}{2 \sqrt{(\frac{1}{a^{2}} + 1)}} \times l n (\frac{\sqrt{\frac{1}{a^{2}} + 1} + p}{\sqrt{\frac{1}{a^{2}} + 1} - p})$ $= \frac{1}{2 \sqrt{1 + a^{2}}} l n (\frac{\sqrt{\frac{1}{a^{2}} + 1} + \sqrt{\frac{1}{a^{2}} - k}}{\sqrt{\frac{1}{a^{2}} + 1} - \sqrt{\frac{1}{a^{2}} - k}})$ $= \frac{1}{2 \sqrt{1 + a^{2}}} l n (\frac{\sqrt{1 + a^{2}} + \sqrt{1 - a^{2} k}}{\sqrt{1 + a^{2}} - \sqrt{1 - a^{2} k}})$ $= \frac{1}{2 \sqrt{1 + a^{2}}} l n (\frac{\sqrt{1 + a^{2}} + \sqrt{1 - a^{2} t^{2}}}{\sqrt{1 + a^{2}} - \sqrt{1 - a^{2} t^{2}}})$ $= \frac{1}{2 \sqrt{1 + a^{2}}} ∣ l n (\frac{\sqrt{1 + a^{2}} + \sqrt{1 - \frac{a^{2}}{x^{2}}}}{\sqrt{1 + a^{2}} - \sqrt{1 - \frac{a^{2}}{x^{2}}}}) ∣_{a}^{\infty}$ $= \frac{1}{2 \sqrt{1 + a^{2}}} [{l n (\frac{\sqrt{1 + a^{2}} + \sqrt{1 - 0}}{\sqrt{1 + a^{2}} - \sqrt{1 - 0}})} - l n (\frac{\sqrt{1 + a^{2}} + \sqrt{0}}{\sqrt{1 + a^{2}} - \sqrt{0}})}]$ $= \frac{1}{\sqrt{1 + a^{2}}} l n (\frac{\sqrt{1 + a^{2}} + 1}{\sqrt{1 + a^{2}} - 1})$
	Commented byAbdo msup. last updated on 17/Dec/18

	$t h a n k y o u s i r t a n m a y .$
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