Question Number 121500 by bramlexs22 last updated on 08/Nov/20
![(1) ∫ (dx/(x^4 +2x^2 +9)) ? (2) arc tan ((x/3))−arc tan ((x/3))= arc tan ((1/5))](https://www.tinkutara.com/question/Q121500.png)
Commented by benjo_mathlover last updated on 08/Nov/20
![i think it should be (2) tan^(−1) ((x/3))−tan^(−1) ((x/2))= tan^(−1) ((1/5)) let x = 6ψ ⇒ tan^(−1) (2ψ)−tan^(−1) (3ψ)=tan^(−1) ((1/5)) ⇔ ((2ψ−3ψ)/(1+6ψ^2 )) = (1/5) ; 6ψ^2 +5ψ+1=0 (2ψ+1)(3ψ+1)=0 → { (((1/6)x=−(1/2);x =−3)),(((1/6)x=−(1/3); x=−2)) :}](https://www.tinkutara.com/question/Q121505.png)
Answered by Ar Brandon last updated on 09/Nov/20
![1\I=∫(dx/(x^4 +2x^2 +9))=(1/6)∫(((x^2 +3)−(x^2 −3))/(x^4 +2x^2 +9))dx 6I=∫((x^2 +3)/(x^4 +2x^2 +9))dx−∫((x^2 −3)/(x^4 +2x^2 +9))dx =∫((1+(3/x^2 ))/(x^2 +2+(9/x^2 )))dx−∫((1−(3/x^2 ))/(x^2 +2+(9/x^2 )))dx =∫((1+(3/x^2 ))/((x−(3/x))^2 +8))dx−∫((1−(3/x^2 ))/((x+(3/x))^2 −4))dx =∫(du/(u^2 +8))−∫(dv/(v^2 −4))=(1/(2(√2)))Arctan((u/(2(√2))))+(1/2)Arctanh((v/2))+C](https://www.tinkutara.com/question/Q121507.png)
Answered by MJS_new last updated on 09/Nov/20
![∫(dx/(x^4 +2x^2 +9))=∫(dx/((x^2 −2x+3)(x^2 +2x+3)))= =(1/(12))∫(((x+2)/(x^2 +2x+3))−((x−2)/(x^2 −2x+3)))dx= =(1/(12))∫(dx/(x^2 +2x+3))+(1/(24))∫((2x+2)/(x^2 +2x+3))dx+ +(1/(12))∫(dx/(x^2 −2x+3))−(1/(24))∫((2x−2)/(x^2 −2x+3))dx= =((√2)/(24))arctan (((√2)(x+1))/2) +(1/(24))ln (x^2 +2x+3) + +((√2)/(24))arctan (((√2)(x−1))/2) −(1/(24))ln (x^2 −2x+3) +C](https://www.tinkutara.com/question/Q121508.png)
Answered by TANMAY PANACEA last updated on 09/Nov/20
![1)∫((1/x^2 )/(x^2 +(9/x^2 )+2))dx (1/6)∫(((1+(3/x^2 ))−(1−(3/x^2 )))/(x^2 +(9/x^2 )+2))dx (1/6)∫((d(x−(3/x)))/((x−(3/(x )))^2 +8)) −(1/6)∫((d(x+(3/x)))/((x+(3/x))^2 −4)) (1/6)×(1/( (√8)))tan^(−1) (((x−(3/x))/( (√8))))−(1/6)×(1/(2×2))ln(((x−(3/x)−2)/(x−(3/x)+2)))+c](https://www.tinkutara.com/question/Q121533.png)