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Question Number 106555 by  M±th+et+s last updated on 06/Aug/20

∫_0 ^4 ∫_(√y) ^2 (1/(1+x^3 ))dx dy

04y211+x3dxdy

Commented by bobhans last updated on 06/Aug/20

=syntax error=  ∫_0 ^4 ∫_(√y) ^2  (1/(1+x^3 )) dydx

=syntaxerror=402y11+x3dydx

Answered by mathmax by abdo last updated on 06/Aug/20

I =∫_0 ^4  (∫_(√y) ^2  (dx/(1+x^3 )))dy =∫_0 ^4  A(y)dy with A(y) =∫_(√y) ^(2 )  (dx/(x^3  +1))  let decompose f(x) =(1/(x^3 +1)) =(1/((x+1)(x^2 −x+1)))  f(x) =(a/(x+1)) +((bx +c)/(x^2 −x+1))  we get a=(1/3)  lim_(x→+∞) xf(x) =0 =a+b ⇒b =−(1/3) ⇒f(x)=(1/(3(x+1)))+((−(1/3)x+c)/(x^2 −x+1))  f(0) =1 =(1/3) +c ⇒c =(2/3) ⇒f(x) =(1/(3(x+1)))+((−(1/3)x+(2/3))/(x^2 −x+1))  ⇒∫ f(x) dx =(1/3)ln∣x+1∣−(1/3)∫ ((x−2)/(x^2 −x+1))dx  but  ∫ ((x−2)/(x^2 −x+1))dx =(1/2)∫ ((2x−1−3)/(x^2 −x+1))dx =(1/2)ln(x^2 −x+1)−(3/2)∫ (dx/(x^2 −x+1))  ∫ (dx/(x^2 −x+1))dx =∫  (dx/(x^2 −2(1/2)x +(1/4)+(3/4))) =∫ (dx/((x−(1/2))^2  +(3/4)))  =_(x−(1/2)=((√3)/2)u)    (4/3)  ∫   (1/(u^2  +1)).((√3)/2)du =(2/(√3)) arctan(((2x−1)/(√3))) +c ⇒  ∫ ((x−2)/(x^2 −x+1))dx =(1/2)ln(x^2 −x+1)−(√3)arctan(((2x−1)/(√3))) ⇒  ∫ f(x)dx =(1/3)ln∣x+1∣−(1/6)ln(x^2 −x+1)+((√3)/3) arctan(((2x−1)/(√3))) ⇒  A(y) =[(1/3)ln∣x+1∣−(1/6)ln(x^2 −x+1)+((√3)/3) arctan(((2x−1)/(√3)))]_(√y) ^2   =(1/3)ln(3)−(1/6)ln(3)+((√3)/3) arctan((√3))−(1/3)ln((√y)+1)  +(1/6)ln(y−(√y)+1)−((√3)/3) arctan(((2(√y)−1)/(√3))) ⇒  I =∫_0 ^4  A(y)dy =4{((ln3)/6) +((√3)/3) arctan((√3))}−(1/3)∫_0 ^4 ln(1+(√y))dy  +(1/6)∫_0 ^4 ln(y−(√y)+1)dy −((√3)/3) ∫_0 ^4  arctan(((2(√y)−1)/(√3)))  we have ∫_0 ^4 ln(1+(√y))dy =_(by parts)   [yln(1+(√y))]_0 ^4 −∫_0 ^4 y.(1/(2(√y)(1+(√y))))dy  =4ln(3)−(1/2)∫_0 ^4  ((√y)/(1+(√y))) dy  (y=t^2 )  =4ln(3)−(1/2) ∫_0 ^2  (t/(1+t))(2t)dt =4ln(3)−∫_0 ^2  (t^2 /(1+t))dt and  ∫_0 ^2  (t^2 /(1+t))dt =∫_0 ^2  ((t^2 −1 +1)/(t+1))dt =∫_0 ^2 (t−1)dt +[ln(t+1)]_0 ^2   =[(t^2 /2)−t]_0 ^2  +ln(3) =0+ln(3) =ln(3)  ∫_0 ^4 ln(y−(√y)+1)dy =_((√y)=t)    ∫_0 ^2 ln(t^2 −t+1)(2t)dt  =2 ∫_0 ^2 t ln(t^2 −t+1)dt =_(by parts)  2{ [(t^2 /2)ln(t^2 −t+1)]_0 ^2 −∫_0 ^2 (t^2 /2).((2t−1)/(t^2 −t+1))dt}  =2{ 2ln(3)−(1/2) ∫_0 ^2  ((2t^3 −t^2 )/(t^2 −t+1))dt}=....(eazy to solve)  also the integral∫_0 ^4  arctan(((2(√y)−1)/(√3))) can be solved by parts  ...be continued...

I=04(y2dx1+x3)dy=04A(y)dywithA(y)=y2dxx3+1letdecomposef(x)=1x3+1=1(x+1)(x2x+1)f(x)=ax+1+bx+cx2x+1wegeta=13limx+xf(x)=0=a+bb=13f(x)=13(x+1)+13x+cx2x+1f(0)=1=13+cc=23f(x)=13(x+1)+13x+23x2x+1f(x)dx=13lnx+113x2x2x+1dxbutx2x2x+1dx=122x13x2x+1dx=12ln(x2x+1)32dxx2x+1dxx2x+1dx=dxx2212x+14+34=dx(x12)2+34=x12=32u431u2+1.32du=23arctan(2x13)+cx2x2x+1dx=12ln(x2x+1)3arctan(2x13)f(x)dx=13lnx+116ln(x2x+1)+33arctan(2x13)A(y)=[13lnx+116ln(x2x+1)+33arctan(2x13)]y2=13ln(3)16ln(3)+33arctan(3)13ln(y+1)+16ln(yy+1)33arctan(2y13)I=04A(y)dy=4{ln36+33arctan(3)}1304ln(1+y)dy+1604ln(yy+1)dy3304arctan(2y13)wehave04ln(1+y)dy=byparts[yln(1+y)]0404y.12y(1+y)dy=4ln(3)1204y1+ydy(y=t2)=4ln(3)1202t1+t(2t)dt=4ln(3)02t21+tdtand02t21+tdt=02t21+1t+1dt=02(t1)dt+[ln(t+1)]02=[t22t]02+ln(3)=0+ln(3)=ln(3)04ln(yy+1)dy=y=t02ln(t2t+1)(2t)dt=202tln(t2t+1)dt=byparts2{[t22ln(t2t+1)]0202t22.2t1t2t+1dt}=2{2ln(3)12022t3t2t2t+1dt}=....(eazytosolve)alsotheintegral04arctan(2y13)canbesolvedbyparts...becontinued...

Commented by Ar Brandon last updated on 06/Aug/20

May I proceed...    ∫_0 ^4 arctan(((2(√y)−1)/(√3)))dy  u(y)=arctan(((2(√y)−1)/(√3)))⇒u′(y)=(1/(√(3y)))∙(1/(1+(((2(√y)−1)/(√3)))^2 ))  ⇒u′(y)=((√3)/(4(√y)(y−(√y)+1)))  v′(y)=1⇒v(y)=y  ⇒I=[y∙arctan(((2(√y)−1)/(√3)))]_0 ^4 −((√3)/4)∫_0 ^4 (y/((√y)(y−(√y)+1)))dy          =((4π)/3)−((√3)/4)∫_0 ^4 ((√y)/((y−(√y)+1)))dy=((4π)/3)−((√3)/4)∫_0 ^2 ((2t^2 )/(t^2 −t+1))dt          =((4π)/3)−((√3)/4)∫_0 ^2 {1+(1/2)∙((t−1)/(t^2 −t+1))}dt          =((4π)/3)−((√3)/2)−((√3)/8)∫_0 ^2 {(1/2)∙((2t−1)/(t^2 −t+1))−(1/2)∙(1/(t^2 −t+1))}dt          =((4π)/3)−((√3)/2)−[((√3)/(16))ln(t^2 −t+1)]_0 ^2 +((√3)/(16))∫_0 ^2 (dt/((t−(1/2))^2 +(3/4)))          =((4π)/3)−((√3)/2)−(((√3)ln3)/(16))+(1/8)[arctan(((2t−1)/(√3)))]_0 ^2           =((4π)/3)−((√3)/2)−(((√3)ln3)/(16))+(1/8)((π/3)+(π/6))

MayIproceed...04arctan(2y13)dyu(y)=arctan(2y13)u(y)=13y11+(2y13)2u(y)=34y(yy+1)v(y)=1v(y)=yI=[yarctan(2y13)]043404yy(yy+1)dy=4π33404y(yy+1)dy=4π334022t2t2t+1dt=4π33402{1+12t1t2t+1}dt=4π3323802{122t1t2t+1121t2t+1}dt=4π332[316ln(t2t+1)]02+31602dt(t12)2+34=4π3323ln316+18[arctan(2t13)]02=4π3323ln316+18(π3+π6)

Answered by 1549442205PVT last updated on 06/Aug/20

  (1/(1+x^3 ))=(1/((x+1)(x^2 −x+1)))=(a/(x+1))+((bx+c)/(x^2 −x+1))  ⇔(a+b)x^2 +(b−a+c)x+a+c=1  ⇔ { ((a+b=0)),((b−a+c=0)),((a+c=1)) :}⇒ { ((2b+c=0)),((−b+c=1)) :}⇒ { ((b=−1/3)),((c=2/3)),((a=1/3)) :}  Hence ∫_(√y) ^( 2) (1/(1+x^3 ))=∫_(√y) ^( 2)  (dx/(3(x+1)))−∫_(√y) ^( 2)  ((x−2)/(3(x^2 −x+1)))dx  =(1/3)ln(x+1)∣_(√y) ^2 −(1/3)((1/2)∫_(√y) ^( 2)  ((2x−1−3)/(x^2 −x+1)))  =(1/3)(ln3−ln(1+(√y)))−(1/6)∫_(√y) ^( 2)  ((d(x^2 −x+1))/((x^2 −x+1)))dx  +(1/2)∫_(√y) ^( 2)  (dx/(x^2 −x+1))=(1/3)(ln3−ln(1+(√y)))  −(1/6)ln(x^2 −x+1)∣_(√y) ^2 +(1/2)×∫ (dx/((x−(1/2))^2 +(((√3)/2))^2 ))  ==(1/3)(ln3−ln(1+(√y)))−(1/6)(ln3−ln(y−(√y)+1)  +(1/2)×(2/(√3))tan^(−1) (((2x−1)/(√3)))∣_(√y) ^2   =(1/6)ln3+(1/6)ln((y−(√y)+1)/(y+2(√y)+1))+(1/(√3))[(π/3)−tan^(−1) (((2(√y)−1)/(√3)))]  Now we calculate  I=∫ _0^4 {(1/6)ln3+(1/6)ln((y−(√y)+1)/(y+2(√y)+1))+(1/(√3))[(π/3)−tan^(−1) (((2(√y)−1)/(√3)))]}dy  =4((1/6)ln3+(π/(3(√3))))+(1/6)∫_0 ^4 ln((y−(√y)+1)/(y+2(√y)+1))dy  −(1/(√3))∫_0 ^( 4) tan^(−1) (((2(√y)−1)/(√3)))dy  Now we find A=∫_0 ^4 ln(y−(√y)+1)dy=  yln(y−(√y)+1)∣_0 ^4 −∫_0 ^4 y(ln(y−(√y)+1)′dy  =4ln3−B  B=∫_0 ^4  y.((1−(1/(2(√y))))/(y−(√y)+1))dy=∫_0 ^( 4) ((y(2(√y)−1))/(2(√y)(y−(√y)+1)))dy  =(1/2)∫_0 ^( 4) ((2y−(√y))/(y−(√y)+1))dy.Set (√y)=u⇒du=(dy/(2(√y)))=(dy/(2u))  ⇒2udu=dy⇒B=(1/2)∫_0 ^( 2)  (((2u^2 −u)2udu)/(u^2 −u+1))  =∫_0 ^( 2)  ((2u^3 −u^2 )/(u^2 −u+1))du=∫_0 ^( 2) (2u+((u−2)/(u^2 −u+1)))du  =u^2 ∣_0 ^2 +(1/2)∫_0 ^( 2)   ((2u−1−3)/(u^2 −u+1))du=u^2 ∣_0 ^2 +(1/2)∫_0 ^( 2)  ((d(u^2 −u+1))/(u^2 −u+1))−(3/2)∫_0 ^( 2)  (du/((u−(1/2))^2 +(((√3)/2))^2 ))  =[u^2 +(1/2)ln(u^2 −u+1)]∣_0 ^2 −(3/2)×(2/(√3))tan^(−1) (((2u−1)/(√3)))∣_0 ^2   =4+(1/2)ln3−(√3) tan^(−1) ((√3))+tan^(−1) (−(1/(√3)))  ⇒A=4ln3−B=3.5ln3−4+((π(√3))/3)+(π/6)  C=∫ _0^4  ln((√y)+1)^2 dy=2∫_0 ^( 4) ln((√y)+1)dy  =2yln((√y)+1)∣_0 ^4 −2∫_0 ^( 4) y.[ln((√y)+1)]′dy  =2yln((√y)+1)∣_0 ^4 −D=8ln3−D  D=2∫_0 ^( 4)  ((y.(1/(2(√y))))/((√y)+1))=∫_0 ^4 ((√y)/((√y)+1))dy=∫_0 ^( 2)  ((2u^2 du)/(u+1))  =2∫_0 ^2 (u^2 −1)du+2∫_0 ^2  (du/(u+1))=[2((u^3 /3)−u)+2ln(u+1)]_0 ^2   =2((8/3)−2)+2ln3⇒C=8ln3−D=6ln3−(4/3)  Next we calculate E= (1/(√3))∫_0 ^( 4) tan^(−1) (((2(√y)−1)/(√3)))dy  Set F=∫_0 ^( 4) tan^(−1) (((2(√y)−1)/(√3))) dy.Putting  ((2(√y)−1)/(√3))=u⇒(√y)=(((√3)u+1)/2)⇒y=(3u^2 +2(√3)u+1)/4  ⇒dy=(1/4)(6u+2(√3))du  F=∫_0 ^( 4) tan^(−1) (u)×(1/4)(6u+2(√3))du  =(3/2)∫_0 ^( 4)  utan^(−1) (u)du+((√3)/2)∫_0 ^( 4) tan^(−1) (u)du  =(3/4)u^2 tan^(−1) (u)−(3/4)∫_0 ^( 4)  ((u^2 du)/(1+u^2 ))+((√3)/2)utan^(−1) (u)−((√3)/2)∫_0 ^( 4)  ((udu)/(1+u^2 ))  =(3/4)u^2 tan^(−1) (u)−(3/4)∫_0 ^( 4) (1−(1/(1+u^2 )))du  +((√3)/2)utan^(−1) (u)−((√3)/4)∫_0 ^( 4) (1/(1+u^2 ))d(u^2 +1)  =[(3/4)u^2 tan^(−1) (u)−(3/4)u+(3/4)tan^(−1) (u)  +((√3)/2)utan^(−1) (u)−((√3)/4)ln(u^2 +1)]∣_(−(1/(√3))) ^(√3)   =(9/4).(π/3)−((3(√3))/4)+(3/4).(π/3)+(3/2).(π/3)−((√3)/4)ln4  −((1/4).((−π)/6)+((√3)/4)−(3/4).(π/6)+(1/2).(π/6)−((√3)/4)ln(4/3))  =((19π)/(12))−(√3)−((√3)/4)ln3⇒E=(1/(√3))F=((19π(√3))/(36))−1−(1/4)ln3  =(1/2)ln3+(((1−(√3))π)/(36))

11+x3=1(x+1)(x2x+1)=ax+1+bx+cx2x+1(a+b)x2+(ba+c)x+a+c=1{a+b=0ba+c=0a+c=1{2b+c=0b+c=1{b=1/3c=2/3a=1/3Hencey211+x3=y2dx3(x+1)y2x23(x2x+1)dx=13ln(x+1)y213(12y22x13x2x+1)=13(ln3ln(1+y))16y2d(x2x+1)(x2x+1)dx+12y2dxx2x+1=13(ln3ln(1+y))16ln(x2x+1)y2+12×dx(x12)2+(32)2==13(ln3ln(1+y))16(ln3ln(yy+1)+12×23tan1(2x13)y2=16ln3+16lnyy+1y+2y+1+13[π3tan1(2y13)]NowwecalculateI=04{16ln3+16lnyy+1y+2y+1+13[π3tan1(2y13)]}dy=4(16ln3+π33)+1604lnyy+1y+2y+1dy1304tan1(2y13)dyNowwefindA=04ln(yy+1)dy=yln(yy+1)0404y(ln(yy+1)dy=4ln3BB=04y.112yyy+1dy=04y(2y1)2y(yy+1)dy=12042yyyy+1dy.Sety=udu=dy2y=dy2u2udu=dyB=1202(2u2u)2uduu2u+1=022u3u2u2u+1du=02(2u+u2u2u+1)du=u202+12022u13u2u+1du=u202+1202d(u2u+1)u2u+13202du(u12)2+(32)2=[u2+12ln(u2u+1)]0232×23tan1(2u13)02=4+12ln33tan1(3)+tan1(13)A=4ln3B=3.5ln34+π33+π6C=04ln(y+1)2dy=204ln(y+1)dy=2yln(y+1)04204y.[ln(y+1)]dy=2yln(y+1)04D=8ln3DD=204y.12yy+1=04yy+1dy=022u2duu+1=202(u21)du+202duu+1=[2(u33u)+2ln(u+1)]02=2(832)+2ln3C=8ln3D=6ln343NextwecalculateE=1304tan1(2y13)dySetF=04tan1(2y13)dy.Putting2y13=uy=3u+12y=(3u2+23u+1)/4dy=14(6u+23)duF=04tan1(u)×14(6u+23)du=3204utan1(u)du+3204tan1(u)du=34u2tan1(u)3404u2du1+u2+32utan1(u)3204udu1+u2=34u2tan1(u)3404(111+u2)du+32utan1(u)340411+u2d(u2+1)=[34u2tan1(u)34u+34tan1(u)+32utan1(u)34ln(u2+1)]133=94.π3334+34.π3+32.π334ln4(14.π6+3434.π6+12.π634ln43)=19π12334ln3E=13F=19π336114ln3=12ln3+(13)π36

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