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Evaluate-1-ax-2-bx-c-dx-




Question Number 83807 by niroj last updated on 06/Mar/20
  Evaluate:    ∫  (( 1)/(ax^2 +bx+c))dx
Evaluate:1ax2+bx+cdx
Commented by mathmax by abdo last updated on 06/Mar/20
I =∫  (dx/(ax^2  +bx +c))  case 1  a=0 and b≠0 ⇒I =(1/b)ln∣bx+c∣ +K now we take a≠0  Δ =b^2 −4ac   case 1  Δ>0 ⇒x_1 =((−b+(√(b^2 −ac)))/(2a))  and x_2 =((−b−(√(b^2 −4ac)))/(2a))  I =∫  (dx/(a(x−x_1 )(x−x_2 ))) =(1/a)∫ ((1/(x−x_1 ))−(1/(x−x_2 )))dx  =(1/( (√(b^2 −4ac))))ln∣((x−x_1 )/(x−x_2 ))∣ +C  case2  Δ<0 ⇒b^2 <4ac ⇒I =∫  (dx/(a( x^2  +(b/a)x +(c/a))))  =(1/a)∫  (dx/((x^2  +2(b/(2a))x +(b^2 /(4a^2 )) +(c/a)−(b^2 /(4a^2 )))))  =(1/a) ∫   (dx/((x+(b/(2a)))^2  +((4ac−b^2 )/(4a^2 ))))  =_(x+(b/(2a))=(√((4ac−b^2 )/(4a^2 )))z) (1/a)∫  (1/(((4ac−b^2 )/(4a^2 ))(1+z^2 )))×(√((4ac−b^2 )/(4a^2 ))) dz  =((4a)/(4ac−b^2 ))×((√(4ac−b^2 ))/(2∣a∣)) arctan((1/( (√((4ac−b^2 )/(4a^2 )))))(x+(b/(2a)))) +K  =((2a)/(∣a∣(√(4ac−b^2 )))) arctan((1/( (√((4ac−b^2 )/(4a^2 )))))(x+(b/(2a)))) +K  =
I=dxax2+bx+ccase1a=0andb0I=1blnbx+c+Knowwetakea0Δ=b24accase1Δ>0x1=b+b2ac2aandx2=bb24ac2aI=dxa(xx1)(xx2)=1a(1xx11xx2)dx=1b24aclnxx1xx2+Ccase2Δ<0b2<4acI=dxa(x2+bax+ca)=1adx(x2+2b2ax+b24a2+cab24a2)=1adx(x+b2a)2+4acb24a2=x+b2a=4acb24a2z1a14acb24a2(1+z2)×4acb24a2dz=4a4acb2×4acb22aarctan(14acb24a2(x+b2a))+K=2aa4acb2arctan(14acb24a2(x+b2a))+K=
Answered by john santu last updated on 06/Mar/20
(1/a) ∫ (dx/(x^2 +(b/a)x+(c/a))) =   (1/a) ∫ (dx/((x+(b/(2a)))^2 +(c/a)−(b^2 /(4a^2 )))) =  (1/a) ∫ (dx/((x+(b/(2a)))^2 +((√((4ac−b^2 )/(4a^2 ))))^2 )) =  ((2a)/( (√(4ac−b^2 )))) tan^(−1) (((4a^2 x+2ab)/( (√(4ac−b^2 ))))) + c
1adxx2+bax+ca=1adx(x+b2a)2+cab24a2=1adx(x+b2a)2+(4acb24a2)2=2a4acb2tan1(4a2x+2ab4acb2)+c
Answered by TANMAY PANACEA last updated on 06/Mar/20
ax^2 +bx+c  (1/(4a))(4a^2 x^2 +4abx+4ac)  (1/(4a)){(2ax+b)^2 +4ac−b^2 }  condition...(1)  when b^2 >4ac  (1/(4a)){(2ax+b)^2 −(b^2 −4ac)}  (1/(4a)){(2ax+b)^2 −((√(b^2 −4ac)) )^2 }  so ∫(dx/(ax^2 +bx+c))  ∫(dx/((1/(4a)){(2ax+b)^2 −((√(b^2 −4ac)) )^2 ))  now pls use formula∫(dy/(y^2 −k^2 ))  condition 2  b^2 <4ac  ∫(dx/((1/(4a)){(2ax+b)^2 +((√(4ac−b^2 )) )^2 ))  now use formula ∫(dy/(y^2 +k^2 ))  condition 3  b^2 =4ac  ∫(dx/((1/(4a))(2ax+b)^2 ))
ax2+bx+c14a(4a2x2+4abx+4ac)14a{(2ax+b)2+4acb2}condition(1)whenb2>4ac14a{(2ax+b)2(b24ac)}14a{(2ax+b)2(b24ac)2}sodxax2+bx+cdx14a{(2ax+b)2(b24ac)2nowplsuseformuladyy2k2condition2b2<4acdx14a{(2ax+b)2+(4acb2)2nowuseformuladyy2+k2condition3b2=4acdx14a(2ax+b)2
Commented by niroj last updated on 06/Mar/20
thanks to all.
thankstoall.

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