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Question Number 47916 by peter frank last updated on 16/Nov/18

Commented by maxmathsup by imad last updated on 17/Nov/18

$b) l e t I = \int \frac{\sqrt{x^{2} + 2 x - 3}}{x + 2} d x \Rightarrow I = \int \frac{\sqrt{{(x + 1)}^{2} - 4}}{x + 2} d x =_{x + 1 = 2 c h t} \int \frac{2 \sqrt{{c h}^{2} t - 1}}{2 c h (t) - 1 + 2} 2 s h (t) d t$ $= 4 \int \frac{{s h}^{2} t}{2 c h (t) + 1} d t = 4 \int \frac{\frac{c h (2 t) - 1}{2}}{2 c h (t) + 1} d t = 2 \int \frac{c h (2 t) - 1}{2 c h (t) + 1} d t$ $= 2 \int \frac{\frac{e^{2 t} + e^{- 2 t}}{2} - 1}{2 \frac{e^{t} + e^{- t}}{2} + 1} d t = \int \frac{e^{2 t} + e^{- 2 t} - 2}{e^{t} + e^{- t} + 1} d t$ $=_{e^{t} = u} \int \frac{u^{2} + u^{- 2} - 2}{u + u^{- 1} + 1} d u = \int \frac{u^{4} + 1 - 2 u^{2}}{u^{3} + u + u^{2}} d u$ $= \int \frac{u^{4} - 2 u^{2} + 1}{u^{3} + u^{2} + u} d u = \int \frac{u (u^{3} + u^{2} + u) - u^{3} - u^{2} - 2 u^{2} + 1}{u^{3} + u^{2} + u} d u$ $= \frac{u^{2}}{2} - \int \frac{u^{3} + 3 u^{2} - 1}{u^{3} + u^{2} + u} d u = \frac{u^{2}}{2} - \int \frac{u^{3} + u^{2} + u - u^{2} - u + 3 u^{2} - 1}{u^{3} + u^{2} + u} d u$ $= \frac{u^{2}}{2} - u - \int \frac{2 u^{2} - u - 1}{u^{3} + u^{2} + u} d u l e t d e c o m p o s e F (u) = \frac{2 u^{2} - u - 1}{u^{3} + u^{2} + u}$ $F (u) = \frac{a}{u} + \frac{b u + c}{u^{2} + u + 1}$ $a = {l i m}_{u \to 0} u F (u) = - \frac{1}{3}$ ${l i m}_{u \to + \infty} u F (u) = 2 = a + b \Rightarrow b = 2 + \frac{1}{3} = \frac{7}{3} \Rightarrow F (u) = - \frac{1}{3 u} + \frac{\frac{7}{3} u + c}{u^{2} + u + 1}$ $F (1) = 0 = - \frac{1}{3} + \frac{1}{3} (\frac{7}{3} + c) \Rightarrow - 1 + \frac{7}{3} + c = 0 \Rightarrow c = - \frac{4}{3} \Rightarrow$ $F (u) = - \frac{1}{3 u} + \frac{1}{3} \frac{7 u - 4}{u^{2} + u + 1} . . . b e c o n t i n u e d . . .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 16/Nov/18
	$b)∫((x^2 +2x−3)/((x+2)(√(x^2 +2x−3)) ))dx ∫(x/(√(x^2 +2x−3)))−3∫(dx/((x+2)(√(x^2 +2x−3)))) (1/2)∫((2x+2−2)/(√(x^2 +2x−3)))−3∫(dx/((x+2)(√(x^2 +2x−3)))) (1/2)∫((d(x^2 +2x−3))/((√(x^2 +2x−3)) ))dx−∫(dx/(√((x+1)^2 −2^2 )))−3∫(dx/((x+2)(√(x^2 +2x−3)))) I_1 =(1/2)∫((d(x^2 +2x−3))/(√(x^2 +2x−3))) =(1/2)×((√(x^2 +2x−3))/(1/2))=(√(x^2 +2x−3)) +c_1 I_2 =∫(dx/((√((x+1)^2 −2^2 )) )) =ln{(x+1)+(√((x+1)^2 −2^2 )) =ln{(x+1)+(√(x^2 +2x−3)) }+c_2 I_3 =∫(dx/((x+2)(√(x^2 +2x−3)))) t=(1/(x+2)) x+2=(1/t) dx=((−1)/t^2 )dt ∫((−dt)/(t^2 ×(1/t)×(√(((1/t)−2)^2 +2((1/t)−2)−3)))) ∫((−dt)/(t(√((1/t^2 )−(4/t)+4+(2/t)−4−3)))) ∫((−dt)/(t(√((1−4t+4t^2 +2t−7t^2 )/t^2 )))) ∫((−dt)/(√(−3t^2 −2t+1))) ∫((−dt)/(√(1−3(t^2 +(2/3)t+(1/9)−(1/9))))) ∫((−dt)/(√(1−3(t+(1/3))^2 +(1/3)))) ∫((−dt)/(√((4/3)−3(t+(1/3))^2 ))) ((−1)/(√3))∫(dt/(√(((2/3))^2 −(t+(1/3))^2 ))) ((−1)/(√3))×sin^(−1) (((t+(1/3))/(2/3)))+c_3 ((−1)/(√3))sin^(−1) ((((1/(x+2))+(1/3))/(2/3)))+c_3 pls add them...$
	$b) \int \frac{x^{2} + 2 x - 3}{(x + 2) \sqrt{x^{2} + 2 x - 3}} d x$ $\int \frac{x}{\sqrt{x^{2} + 2 x - 3}} - 3 \int \frac{d x}{(x + 2) \sqrt{x^{2} + 2 x - 3}}$ $\frac{1}{2} \int \frac{2 x + 2 - 2}{\sqrt{x^{2} + 2 x - 3}} - 3 \int \frac{d x}{(x + 2) \sqrt{x^{2} + 2 x - 3}}$ $\frac{1}{2} \int \frac{d (x^{2} + 2 x - 3)}{\sqrt{x^{2} + 2 x - 3}} d x - \int \frac{d x}{\sqrt{{(x + 1)}^{2} - 2^{2}}} - 3 \int \frac{d x}{(x + 2) \sqrt{x^{2} + 2 x - 3}}$ $I_{1} = \frac{1}{2} \int \frac{d (x^{2} + 2 x - 3)}{\sqrt{x^{2} + 2 x - 3}} = \frac{1}{2} \times \frac{\sqrt{x^{2} + 2 x - 3}}{\frac{1}{2}} = \sqrt{x^{2} + 2 x - 3} + c_{1}$ $I_{2} = \int \frac{d x}{\sqrt{{(x + 1)}^{2} - 2^{2}}}$ $= l n {(x + 1) + \sqrt{{(x + 1)}^{2} - 2^{2}}$ $= l n {(x + 1) + \sqrt{x^{2} + 2 x - 3}} + c_{2}$ $I_{3} = \int \frac{d x}{(x + 2) \sqrt{x^{2} + 2 x - 3}}$ $t = \frac{1}{x + 2} x + 2 = \frac{1}{t} d x = \frac{- 1}{t^{2}} d t$ $\int \frac{- d t}{t^{2} \times \frac{1}{t} \times \sqrt{{(\frac{1}{t} - 2)}^{2} + 2 (\frac{1}{t} - 2) - 3}}$ $\int \frac{- d t}{t \sqrt{\frac{1}{t^{2}} - \frac{4}{t} + 4 + \frac{2}{t} - 4 - 3}}$ $\int \frac{- d t}{t \sqrt{\frac{1 - 4 t + 4 t^{2} + 2 t - 7 t^{2}}{t^{2}}}}$ $\int \frac{- d t}{\sqrt{- 3 t^{2} - 2 t + 1}}$ $\int \frac{- d t}{\sqrt{1 - 3 (t^{2} + \frac{2}{3} t + \frac{1}{9} - \frac{1}{9})}}$ $\int \frac{- d t}{\sqrt{1 - 3 {(t + \frac{1}{3})}^{2} + \frac{1}{3}}}$ $\int \frac{- d t}{\sqrt{\frac{4}{3} - 3 {(t + \frac{1}{3})}^{2}}}$ $\frac{- 1}{\sqrt{3}} \int \frac{d t}{\sqrt{{(\frac{2}{3})}^{2} - {(t + \frac{1}{3})}^{2}}}$ $\frac{- 1}{\sqrt{3}} \times {s i n}^{- 1} (\frac{t + \frac{1}{3}}{\frac{2}{3}}) + c_{3}$ $\frac{- 1}{\sqrt{3}} {s i n}^{- 1} (\frac{\frac{1}{x + 2} + \frac{1}{3}}{\frac{2}{3}}) + c_{3}$ $p l s a d d t h e m . . .$
	Commented by peter frank last updated on 16/Nov/18

	$thank you sir . . .$
	Commented by tanmay.chaudhury50@gmail.com last updated on 17/Nov/18

	$m o s t w e l c o m e . . .$
	Commented by malwaan last updated on 17/Nov/18

	$wonderful$
	Answered by tanmay.chaudhury50@gmail.com last updated on 17/Nov/18
	$∫cotxcot2xcot3xdx =∫((cosxcos2xcos3x)/(sinxsin2xsin3x))dx =∫((cosx(1−2sin^2 x)(4cos^3 x−3cosx))/(sinx×2sinxcosx×(3sinx−4sin^3 x)))dx =∫(((1−2sin^2 x){cosx(4cos^2 x−3)})/(2sin^2 x(3sinx−4sin^3 x)))dx =∫(((1−2sin^2 x)(4−4sin^2 x−3)cosx)/(2sin^2 x(3sinx−4sin^3 x)))dx t=sinx dt=cosxdx ∫(((1−2t^2 )(1−4t^2 ))/(2t^2 (3t−4t^3 )))dt ∫((1−6t^2 +8t^4 )/(2t^3 (3−4t^2 )))dt=(1/2)∫((1−6t^2 +8t^4 )/(t^3 (3−4t^2 ))) ((1−6t^2 +8t^4 )/(t^3 (3−4t^2 )))=(a/t)+(b/t^2 )+(c/t^3 )+((pt+q)/(3−4t^2 )) 1−6t^2 +8t^4 =at^2 (3−4t^2 )+bt(3−4t^2 )+c(3−4t^2 )+(pt+q)t^3 1−6t^2 +8t^4 =t^2 (3a)+t^4 (−4a)+t(3b)+t^3 (−4b)+3c+t^2 (−4c)+t^4 (p)+t^3 (q) 8t^4 −6t^2 +1=t^4 (−4a+p)+t^3 (−4b+q)+t^2 (3a−4c)+t(3b)+3c −4a+p=8 −4b+q=0 3a−4c=−6 3b=0 3c=1 so b=0 q=0 c=(1/3) 3a=−6+4×(1/3)=((−18+4)/3) a=((−14)/9) p=8+4a =8+((4×−14)/9)=((72−56)/9)=((−16)/9) pls wait busy... ∫(a/t)dt+∫(b/t^2 )dt+∫(c/t^3 )dt+∫((pt+q)/(3−4t^2 ))dt =((−14)/9)∫(dt/t)+(1/3)∫(dt/t^3 )+((−16)/9)∫((tdt)/(3−4t^2 )) =((−14)/9)∫(dt/t)+(1/3)∫t^(−3) dt+(2/9)∫((d(3−4t^2 ))/(3−4t^2 )) =((−14)/9)lnt+(1/3)×(1/(−2t^2 ))+(2/9)ln(3−4t^2 ) so ans is =(1/2)[((−14)/9)ln(sinx)−(1/(6sin^2 x))+(2/9)ln(3−4sin^2 x)]+c pls check...$
	$\int c o t x c o t 2 x c o t 3 x d x$ $= \int \frac{c o s x c o s 2 x c o s 3 x}{s i n x s i n 2 x s i n 3 x} d x$ $= \int \frac{c o s x (1 - 2 {s i n}^{2} x) (4 {c o s}^{3} x - 3 c o s x)}{s i n x \times 2 s i n x c o s x \times (3 s i n x - 4 {s i n}^{3} x)} d x$ $= \int \frac{(1 - 2 {s i n}^{2} x) {c o s x (4 {c o s}^{2} x - 3)}}{2 {s i n}^{2} x (3 s i n x - 4 {s i n}^{3} x)} d x$ $= \int \frac{(1 - 2 {s i n}^{2} x) (4 - 4 {s i n}^{2} x - 3) c o s x}{2 {s i n}^{2} x (3 s i n x - 4 {s i n}^{3} x)} d x$ $t = s i n x d t = c o s x d x$ $\int \frac{(1 - 2 t^{2}) (1 - 4 t^{2})}{2 t^{2} (3 t - 4 t^{3})} d t$ $\int \frac{1 - 6 t^{2} + 8 t^{4}}{2 t^{3} (3 - 4 t^{2})} d t = \frac{1}{2} \int \frac{1 - 6 t^{2} + 8 t^{4}}{t^{3} (3 - 4 t^{2})}$ $\frac{1 - 6 t^{2} + 8 t^{4}}{t^{3} (3 - 4 t^{2})} = \frac{a}{t} + \frac{b}{t^{2}} + \frac{c}{t^{3}} + \frac{p t + q}{3 - 4 t^{2}}$ $1 - 6 t^{2} + 8 t^{4} = {a t}^{2} (3 - 4 t^{2}) + b t (3 - 4 t^{2}) + c (3 - 4 t^{2}) + (p t + q) t^{3}$ $1 - 6 t^{2} + 8 t^{4} = t^{2} (3 a) + t^{4} (- 4 a) + t (3 b) + t^{3} (- 4 b) + 3 c + t^{2} (- 4 c) + t^{4} (p) + t^{3} (q)$ $8 t^{4} - 6 t^{2} + 1 = t^{4} (- 4 a + p) + t^{3} (- 4 b + q) + t^{2} (3 a - 4 c) + t (3 b) + 3 c$ $- 4 a + p = 8$ $- 4 b + q = 0$ $3 a - 4 c = - 6$ $3 b = 0$ $3 c = 1$ $s o b = 0 q = 0 c = \frac{1}{3}$ $3 a = - 6 + 4 \times \frac{1}{3} = \frac{- 18 + 4}{3} a = \frac{- 14}{9}$ $p = 8 + 4 a$ $= 8 + \frac{4 \times - 14}{9} = \frac{72 - 56}{9} = \frac{- 16}{9}$ $p l s w a i t b u s y . . .$ $\int \frac{a}{t} d t + \int \frac{b}{t^{2}} d t + \int \frac{c}{t^{3}} d t + \int \frac{p t + q}{3 - 4 t^{2}} d t$ $= \frac{- 14}{9} \int \frac{d t}{t} + \frac{1}{3} \int \frac{d t}{t^{3}} + \frac{- 16}{9} \int \frac{t d t}{3 - 4 t^{2}}$ $= \frac{- 14}{9} \int \frac{d t}{t} + \frac{1}{3} \int t^{- 3} d t + \frac{2}{9} \int \frac{d (3 - 4 t^{2})}{3 - 4 t^{2}}$ $= \frac{- 14}{9} l n t + \frac{1}{3} \times \frac{1}{- 2 t^{2}} + \frac{2}{9} l n (3 - 4 t^{2})$ $s o a n s i s$ $= \frac{1}{2} [\frac{- 14}{9} l n (s i n x) - \frac{1}{6 {s i n}^{2} x} + \frac{2}{9} l n (3 - 4 {s i n}^{2} x)] + c$ $p l s c h e c k . . .$
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