find ∫ (dx/((x^2 +x)(√(−x^2 +2x +3))))


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Question Number 58220 by maxmathsup by imad last updated on 20/Apr/19

$f i n d \int \frac{d x}{(x^{2} + x) \sqrt{- x^{2} + 2 x + 3}}$
	Answered by tanmay last updated on 20/Apr/19
	$−x^2 −x+3x+3 −x(x+1)+3(x+1) (x+1)(3−x) t^2 =x+1 dx=2tdt ∫(dx/(x(x+1)(√((x+1)(3−x))))) ∫((2tdt)/((t^2 −1)t^2 (√(t^2 (3−t^2 +1))))) ∫((2dt)/((t^2 −1)t^2 (√(4−t^2 )))) 2∫((t^2 −(t^2 −1))/(t^2 (t^2 −1)(√(4−t^2 ))))dt 2∫(dt/((t^2 −1)(√(4−t^2 ))))−2∫(dt/(t^2 (√(4−t^2 )) )) 2I_1 −2I_2 I_1 =∫(dt/((t^2 −1)(√(4−t^2 )))) a^2 =((4−t^2 )/(t^2 −1))→a^2 t^2 −a^2 =4−t^2 t^2 (a^2 +1)=a^2 +4 t^2 =((a^2 +4)/(a^2 +1)) 2tdt=(((a^2 +1)(2a)−(a^2 +4)(2a))/((a^2 +1)^2 ))da 2tdt=((2a^3 +2a−2a^3 −8a)/((a^2 +1)^2 ))da tdt=((−3ada)/((a^2 +1)^2 )) ∫((−3ada)/((√((a^2 +4)/(a^2 +1)))×(a^2 +1)^2 ))×(1/((((a^2 +4)/(a^2 +1))−1)(√(4−(((a^2 +4)/(a^2 +1))))))) ∫((−3ada)/((√(a^2 +4)) ))×(((√(a^2 +1)) )/((a^2 +1)^2 ))×(((a^2 +1))/(3(√((4a^2 +4−a^2 −4)/(a^2 +1))))) ∫((−3ada)/(√(a^2 +4)))×(1/(3(√3) a)) ((−1)/(√3))∫(da/(√(a^2 +4)))→{((−1)/(√3))ln(a+(√(a^2 +4)) )} so 2I_1 =((−2)/(√3))ln((√((4−t^2 )/(t^2 −1))) +(√(((4−t^2 )/(t^2 −1))+4)) ) =((−2)/(√3))ln((((√(4−t^2 )) +t(√3))/((√(t^2 −1)) )))←value of I_1 lengthy problem wait for I_2 I_2 =∫(dt/(t^2 (√(4−t^2 )) )) t=2sinθ ∫((2cosθdθ)/(4sin^2 θ×2cosθ)) =(1/2)∫cosec^2 θdθ =((−cotθ)/2)=((−1)/2)((((√(4−t^2 )) )/t))←value of I_2 so complete answdr is 2I_1 −2I_2 2×((−2)/(√3))ln((((√(4−t^2 )) +t(√3))/(√(t^2 −1))))−2×((−1)/2)(((√(4−t^2 ))/t))+c t^2 =x+1 =((−4)/(√3))ln((((√(3−x)) +(√(3(x+1))) )/(√x)))+(((√(3−x))/(√(x+1))))+c pls check...$
	$- x^{2} - x + 3 x + 3$ $- x (x + 1) + 3 (x + 1)$ $(x + 1) (3 - x)$ $t^{2} = x + 1 d x = 2 t d t$ $\int \frac{d x}{x (x + 1) \sqrt{(x + 1) (3 - x)}}$ $\int \frac{2 t d t}{(t^{2} - 1) t^{2} \sqrt{t^{2} (3 - t^{2} + 1)}}$ $\int \frac{2 d t}{(t^{2} - 1) t^{2} \sqrt{4 - t^{2}}}$ $2 \int \frac{t^{2} - (t^{2} - 1)}{t^{2} (t^{2} - 1) \sqrt{4 - t^{2}}} d t$ $2 \int \frac{d t}{(t^{2} - 1) \sqrt{4 - t^{2}}} - 2 \int \frac{d t}{t^{2} \sqrt{4 - t^{2}}}$ $2 I_{1} - 2 I_{2}$ $I_{1} = \int \frac{d t}{(t^{2} - 1) \sqrt{4 - t^{2}}}$ $a^{2} = \frac{4 - t^{2}}{t^{2} - 1} \to a^{2} t^{2} - a^{2} = 4 - t^{2}$ $t^{2} (a^{2} + 1) = a^{2} + 4$ $t^{2} = \frac{a^{2} + 4}{a^{2} + 1}$ $2 t d t = \frac{(a^{2} + 1) (2 a) - (a^{2} + 4) (2 a)}{{(a^{2} + 1)}^{2}} d a$ $2 t d t = \frac{2 a^{3} + 2 a - 2 a^{3} - 8 a}{{(a^{2} + 1)}^{2}} d a$ $t d t = \frac{- 3 a d a}{{(a^{2} + 1)}^{2}}$ $\int \frac{- 3 a d a}{\sqrt{\frac{a^{2} + 4}{a^{2} + 1}} \times {(a^{2} + 1)}^{2}} \times \frac{1}{(\frac{a^{2} + 4}{a^{2} + 1} - 1) \sqrt{4 - (\frac{a^{2} + 4}{a^{2} + 1})}}$ $\int \frac{- 3 a d a}{\sqrt{a^{2} + 4}} \times \frac{\sqrt{a^{2} + 1}}{{(a^{2} + 1)}^{2}} \times \frac{(a^{2} + 1)}{3 \sqrt{\frac{4 a^{2} + 4 - a^{2} - 4}{a^{2} + 1}}}$ $\int \frac{- 3 a d a}{\sqrt{a^{2} + 4}} \times \frac{1}{3 \sqrt{3} a}$ $\frac{- 1}{\sqrt{3}} \int \frac{d a}{\sqrt{a^{2} + 4}} \to {\frac{- 1}{\sqrt{3}} l n (a + \sqrt{a^{2} + 4})}$ $s o 2 I_{1} = \frac{- 2}{\sqrt{3}} l n (\sqrt{\frac{4 - t^{2}}{t^{2} - 1}} + \sqrt{\frac{4 - t^{2}}{t^{2} - 1} + 4})$ $= \frac{- 2}{\sqrt{3}} l n (\frac{\sqrt{4 - t^{2}} + t \sqrt{3}}{\sqrt{t^{2} - 1}}) \leftarrow v a l u e o f I_{1}$ $l e n g t h y p r o b l e m w a i t f o r I_{2}$ $I_{2} = \int \frac{d t}{t^{2} \sqrt{4 - t^{2}}}$ $t = 2 s i n θ$ $\int \frac{2 c o s θ d θ}{4 {s i n}^{2} θ \times 2 c o s θ}$ $= \frac{1}{2} \int {c o s e c}^{2} θ d θ$ $= \frac{- c o t θ}{2} = \frac{- 1}{2} (\frac{\sqrt{4 - t^{2}}}{t}) \leftarrow v a l u e o f I_{2}$ $s o c o m p l e t e a n s w d r i s 2 I_{1} - 2 I_{2}$ $2 \times \frac{- 2}{\sqrt{3}} l n (\frac{\sqrt{4 - t^{2}} + t \sqrt{3}}{\sqrt{t^{2} - 1}}) - 2 \times \frac{- 1}{2} (\frac{\sqrt{4 - t^{2}}}{t}) + c$ $t^{2} = x + 1$ $= \frac{- 4}{\sqrt{3}} l n (\frac{\sqrt{3 - x} + \sqrt{3 (x + 1)}}{\sqrt{x}}) + (\frac{\sqrt{3 - x}}{\sqrt{x + 1}}) + c$ $p l s c h e c k . . .$
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