Question Number 102462 by Ar Brandon last updated on 09/Jul/20

Calculate ;  J=∫(dx/(x(x^2 +x−1)^2 ))  K=∫((x^3 +x−1)/((x^2 +2)^2 ))dx  L=∫(dx/(x+(√(x^2 +1))))

Answered by bemath last updated on 09/Jul/20

L=((x−(√(x^2 +1)))/(x^2 −(x^2 +1)))= (((√(x^2 +1))−x)/1)  ∫ {(√(x^2 +1))−x} dx =  I_1  = ∫(√(x^2 +1)) dx   x= tan p ⇒dx = sec ^2 p dp  I_1 = ∫sec ^3 p dp = ((sin p)/(2cos ^2 p))+(1/2)ln∣tan ((p/2)+(π/4))∣  I_2  = ∫x dx = (1/2)x^2    I=I_1 −I_2

Answered by Dwaipayan Shikari last updated on 09/Jul/20

L=∫((x−(√(x^2 +1)))/(x^2 −x^2 −1))=∫(√(x^2 +1))dx−∫xdx=(x/2)(√(x^2 +1))+(1/2)log(x+(√(x^2 +1)))−(x^2 /2)+Constant

Answered by PRITHWISH SEN 2 last updated on 09/Jul/20

J=∫(dx/x)−∫(((x+1))/(x^2 +x−1))dx +∫(((x+1))/((x^2 +x−1)^2 )) dx      For  K the decomposition leads to  K= (1/4)∫((4x^3 +8x)/((x^2 +2)^2 ))dx−(1/2)∫((2x)/((x^2 +2)^2 )) −∫(dx/((x^2 +2)^2 ))    = (1/4)ln∣x^4 +4x^2 +4∣+(1/(2(x^2 +2)))−(x/(4(x^2 +2)))−(1/(4(√2)))tan^(−1) ((x/(√2)))+C

Commented byAr Brandon last updated on 09/Jul/20

Thanks for your time.👷

Answered by 1549442205 last updated on 09/Jul/20

K=∫((x(x^2 +2)−(x+1))/((x^2 +2)^2 ))dx=∫(x/(x^2 +2))dx−∫((x+1)/((x^2 +2)^2 ))dx  =(1/2)∫((d(x^2 +2))/(x^2 +2))−(1/2)∫((d(x^2 +2))/((x^2 +2)^2 ))−∫(dx/((x^2 +2)^2 ))  K=(1/2)ln(x^2 +2)+(1/(2(x^2 +2)))−(x/(4(x^2 +2)))−(1/4)×(1/(√2))arctan(x/(√2))  =(1/2)ln(x^2 +2)+((2−x)/(4(x^2 +2)))−(1/(4(√2)))arctan(x/(√2))+C

Answered by Dwaipayan Shikari last updated on 09/Jul/20

K=∫((x^3 +x)/(x^4 +2x^2 +4))dx−{∫(1/((x^2 +2)^2 ))dx}=I_a   K=(1/4)∫((4x^3 +4x)/(x^4 +2x^2 +4))−I_(a                        )      (√2)tanθ=x  K=−(1/4).(1/((x^2 +2)))−I_a     {I_a =∫(1/((x^2 +2)^2 ))=(1/4)∫(((√2)sec^2 θdθ)/((tan^2 θ+1)^2 ))        K=−(1/4).(1/((x^2 +2)))−(1/(4(√2)))tan^(−1) (x/(√2))−(1/(8(√2)))sin(2tan^(−1) (x/(√2)))+C  {I_a =(1/(2(√2)))∫cos^2 θdθ=(1/(2(√2)))((θ/2)+(1/4)sin2θ)

Answered by mathmax by abdo last updated on 09/Jul/20

let f(a) =∫  (dx/(x(x^2  +x+a)))  with  a<(1/4)   f^′ (a) =−∫  (dx/(x(x^2 +x+a)^2 ))  and J =−f^′ (−1)  let explicit f(a)  first we decompose F(x) =(1/(x(x^2  +x+a)))  x^2  +x+a =0 →Δ =1−4a >0 ⇒x_1 =((−1+(√(1−4a)))/2)  x_2 =((−1−(√(1−4a)))/2) ⇒F(x) =(1/(x(x−x_1 )(x−x_2 )))=(α/x) +(β/(x−x_1 )) +(λ/(x−x_2 ))    α =(1/a) , β =(1/(x_1 (√(1−4a)))) =(2/(((√(1−4a))−1)(√(1−4a)))))  λ =(1/(x_2 (x_2 −x_1 ))) =(2/((−1−(√(1−4a)))(−(√(1−4a))))) =(2/((1+(√(1−4a)))(√(1−4a)))) ⇒  f(a) =αln∣x∣ +βln∣x−x_1 ∣+λ ln∣x−x_2 ∣ +c  ⇒f(a) =(1/a)ln∣x∣ +(2/(((√(1−4a))−1)(√(1−4a))))ln∣x−((−1+(√(1−4a)))/2)∣  +(2/((1+(√(1−4a)))(√(1−4a))))ln∣x+((1+(√(1−4a)))/2)∣ +c  rest to calculate f^′ (a)   ...be continued...

Answered by mathmax by abdo last updated on 09/Jul/20

L =∫  (dx/(x+(√(1+x^2 ))))  we do the cha7gement  x =sht ⇒  L =∫  ((ch(t))/(sht +ch(t))) dt =∫   (((e^t +e^(−t) )/2)/(((e^t −e^(−t) )/2)+((e^t  +e^(−t) )/2))) dt  =∫  ((e^t  +e^(−t) )/e^t ) dt =∫(1+e^(−2t) )dt =t −(1/2)e^(−2t)  +c  we have t =argsh(x) =ln(x+(√(1+x^2 ))) ⇒e^(2t)  =(x+(√(1+x^2 )))^2  ⇒  L =ln(x+(√(1+x^2 )))−(1/(2(x+(√(1+x^2 )))^2 )) +C

Answered by mathmax by abdo last updated on 13/Jul/20

snother way for J =∫ (dx/(x(x^2 +x−1)^2 ))  x^2  +x−1 =0→Δ =1+4 =5 ⇒α =((−1+(√5))/2) and β =((−1−(√5))/2)  ⇒J =∫  (dx/(x(x−α)^2 (x−β)^2 )) =∫  (dx/(x(((x−α)/(x−β)))^2 (x−β)^4 ))  we do the changement  ((x−α)/(x−β))=t ⇒x−α =tx−βt ⇒(1−t)x =−βt +α ⇒x =((βt−α)/(t−1))   and x−β =((βt−α)/(t−1))−β =((βt−α−βt+β)/(t−1)) =((β−α)/(t−1)) ⇒  (dx/dt) =((β(t−1)−βt +α)/((t−1)^2 )) =((α−β)/((t−1)^2 )) ⇒  J =∫ ((α−β)/((t−1)^2  t^2 (((β−α)/(t−1)))^4 )) dt =(1/((α−β)^3 )) ∫  (((t−1)^4 )/((t−1)^2  t^2 ))dt  =(1/((α−β)^3 )) ∫  (((t−1)^2 )/t^2 ) dt =(1/((α−β)^3 )) ∫ (((t^2 −2t +1)/t^2 ))dt  =(1/((α−β)^3 )) ∫( 1−(2/t) +(1/t^2 ))dt =(1/((α−β)^3 )) { t−2ln∣t∣−(1/t)} +c  =(1/(((√5))^3 )){ ((x−((−1+(√5))/2))/(x+((1+(√5))/2))) −2ln∣((x−((−1+(√5))/2))/(x+((1+(√5))/2)))∣−((x+((1+(√5))/2))/(x−((−1+(√5))/2)))} +c  J=(1/(5(√5))){ ((2x+1−(√5))/(2x+1+(√5))) −2ln∣((2x+1−(√5))/(2x+1+(√5)))∣−((2x+1+(√5))/(2x+1−(√5)))} +c