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Question Number 67907 by A8;15: last updated on 02/Sep/19

Commented by mathmax by abdo last updated on 02/Sep/19
$let I =∫ (dx/(x(√(x^2 +x−6)))) x^2 +x−6=0→Δ=1−4(−6) =25 ⇒x_1 =((−1+5)/2)=2 and x_2 =((−1−5)/2) =−3 ⇒I =∫ (dx/(x(√((x−2)(x+3))))) we do the changement (√(x−2))=t ⇒x−2=t^2 ⇒x=t^2 +2 ⇒I =∫ ((2tdt)/((t^2 +2)t(√(t^2 +2+3)))) =∫ ((2dt)/((t^2 +2)(√(t^2 +5)))) af ter we do the changement t=(√5)sh(u) ⇒ I =∫ ((2(√5)ch(u))/((25sh^2 u +2)(√5)ch(u)))du =∫ ((2du)/(25((ch(2u)−1)/2)+2)) =∫ ((4du)/(25ch(2u)−25+4)) =∫ ((4du)/(25ch(2u)−21)) =∫ ((4du)/(25((e^(2u) +e^(−2u) )/2)−21)) =∫ ((4du)/(25e^(2u) +25e^(−2u) −42)) =_(e^(2u) =t) ∫ (4/(25t+25t^(−1) −42))×(dt/(2t)) =∫ ((2dt)/(25t^2 +25)) =(2/(25)) ∫ (dt/(1+t^2 )) =(2/(25)) arctan(t)+c =(2/(25)) arctan(e^(2u) ) +c but u=argsh((t/(√5))) =ln((t/(√5))+(√(1+(t^2 /5)))) ⇒ e^(2u) =((t/(√5))+(√(1+(t^2 /5))))^2 ⇒ I =(2/(25)) arctan{(((t+(√(5+t^2 )))/(√5)))^2 } +C$
$l e t I = \int \frac{d x}{x \sqrt{x^{2} + x - 6}}$ $x^{2} + x - 6 = 0 \to Δ = 1 - 4 (- 6) = 25 \Rightarrow x_{1} = \frac{- 1 + 5}{2} = 2 a n d$ $x_{2} = \frac{- 1 - 5}{2} = - 3 \Rightarrow I = \int \frac{d x}{x \sqrt{(x - 2) (x + 3)}} w e d o t h e c h a n g e m e n t$ $\sqrt{x - 2} = t \Rightarrow x - 2 = t^{2} \Rightarrow x = t^{2} + 2 \Rightarrow I = \int \frac{2 t d t}{(t^{2} + 2) t \sqrt{t^{2} + 2 + 3}}$ $= \int \frac{2 d t}{(t^{2} + 2) \sqrt{t^{2} + 5}} a f t e r w e d o t h e c h a n g e m e n t t = \sqrt{5} s h (u) \Rightarrow$ $I = \int \frac{2 \sqrt{5} c h (u)}{(25 {s h}^{2} u + 2) \sqrt{5} c h (u)} d u = \int \frac{2 d u}{25 \frac{c h (2 u) - 1}{2} + 2}$ $= \int \frac{4 d u}{25 c h (2 u) - 25 + 4} = \int \frac{4 d u}{25 c h (2 u) - 21}$ $= \int \frac{4 d u}{25 \frac{e^{2 u} + e^{- 2 u}}{2} - 21} = \int \frac{4 d u}{25 e^{2 u} + 25 e^{- 2 u} - 42}$ $=_{e^{2 u} = t} \int \frac{4}{25 t + 25 t^{- 1} - 42} \times \frac{d t}{2 t} = \int \frac{2 d t}{25 t^{2} + 25} = \frac{2}{25} \int \frac{d t}{1 + t^{2}}$ $= \frac{2}{25} a r c t a n (t) + c = \frac{2}{25} a r c t a n (e^{2 u}) + c$ $b u t u = a r g s h (\frac{t}{\sqrt{5}}) = l n (\frac{t}{\sqrt{5}} + \sqrt{1 + \frac{t^{2}}{5}}) \Rightarrow$ $e^{2 u} = {(\frac{t}{\sqrt{5}} + \sqrt{1 + \frac{t^{2}}{5}})}^{2} \Rightarrow I = \frac{2}{25} a r c t a n {{(\frac{t + \sqrt{5 + t^{2}}}{\sqrt{5}})}^{2}} + C$
	Answered by MJS last updated on 02/Sep/19

	$\int \frac{d x}{x \sqrt{x^{2} + x - 6}} =$ $[t = \frac{12 - x}{5 x} \to d x = - \frac{5 x^{2}}{12} d t]$ $= - \frac{\sqrt{6}}{6} \int \frac{d t}{\sqrt{1 - t^{2}}} = - \frac{\sqrt{6}}{6} \arcsin t =$ $= \frac{\sqrt{6}}{6} \arcsin \frac{x - 12}{5 x}$
	Commented by MJS last updated on 02/Sep/19

	$\int \frac{d x}{x \sqrt{(x + a^{2}) (x - b^{2})}} =$ $[t = \frac{(a^{2} - b^{2}) x - 2 a^{2} b^{2}}{(a^{2} + b^{2}) x} \to d x = \frac{(a^{2} + b^{2}) x^{2}}{2 a^{2} b^{2}} d t]$ $= - \frac{1}{\sqrt{a^{2} b^{2}}} \int \frac{d t}{\sqrt{1 - t^{2}}}$
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