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1-dx-x-4-2x-2-9-2-arc-tan-x-3-arc-tan-x-3-arc-tan-1-5-




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))
(1)dxx4+2x2+9?(2)arctan(x3)arctan(x3)=arctan(15)
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)) :}
ithinkitshouldbe(2)tan1(x3)tan1(x2)=tan1(15)letx=6ψtan1(2ψ)tan1(3ψ)=tan1(15)2ψ3ψ1+6ψ2=15;6ψ2+5ψ+1=0(2ψ+1)(3ψ+1)=0{16x=12;x=316x=13;x=2
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
1I=dxx4+2x2+9=16(x2+3)(x23)x4+2x2+9dx6I=x2+3x4+2x2+9dxx23x4+2x2+9dx=1+3x2x2+2+9x2dx13x2x2+2+9x2dx=1+3x2(x3x)2+8dx13x2(x+3x)24dx=duu2+8dvv24=122Arctan(u22)+12Arctanh(v2)+C
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
dxx4+2x2+9=dx(x22x+3)(x2+2x+3)==112(x+2x2+2x+3x2x22x+3)dx==112dxx2+2x+3+1242x+2x2+2x+3dx++112dxx22x+31242x2x22x+3dx==224arctan2(x+1)2+124ln(x2+2x+3)++224arctan2(x1)2124ln(x22x+3)+C
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
1)1x2x2+9x2+2dx16(1+3x2)(13x2)x2+9x2+2dx16d(x3x)(x3x)2+816d(x+3x)(x+3x)2416×18tan1(x3x8)16×12×2ln(x3x2x3x+2)+c

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