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

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

finddx(x2+x)x2+2x+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...

x2x+3x+3x(x+1)+3(x+1)(x+1)(3x)t2=x+1dx=2tdtdxx(x+1)(x+1)(3x)2tdt(t21)t2t2(3t2+1)2dt(t21)t24t22t2(t21)t2(t21)4t2dt2dt(t21)4t22dtt24t22I12I2I1=dt(t21)4t2a2=4t2t21a2t2a2=4t2t2(a2+1)=a2+4t2=a2+4a2+12tdt=(a2+1)(2a)(a2+4)(2a)(a2+1)2da2tdt=2a3+2a2a38a(a2+1)2datdt=3ada(a2+1)23adaa2+4a2+1×(a2+1)2×1(a2+4a2+11)4(a2+4a2+1)3adaa2+4×a2+1(a2+1)2×(a2+1)34a2+4a24a2+13adaa2+4×133a13daa2+4{13ln(a+a2+4)}so2I1=23ln(4t2t21+4t2t21+4)=23ln(4t2+t3t21)valueofI1lengthyproblemwaitforI2I2=dtt24t2t=2sinθ2cosθdθ4sin2θ×2cosθ=12cosec2θdθ=cotθ2=12(4t2t)valueofI2socompleteanswdris2I12I22×23ln(4t2+t3t21)2×12(4t2t)+ct2=x+1=43ln(3x+3(x+1)x)+(3xx+1)+cplscheck...

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