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Question Number 81336 by mathmax by abdo last updated on 11/Feb/20
decompose F(x)=(1/((x−1)^3 (x+3)^7 )) and detrmine ∫ F(x)dx 2) calculate ∫_2 ^(+∞) F(x)dx
decomposeF(x)=1(x−1)3(x+3)7anddetrmine∫F(x)dx2)calculate∫2+∞F(x)dx
Commented by Tony Lin last updated on 12/Feb/20
∫(dx/((x−1)^3 (x+3)^7 )) let u=((x+3)/(x−1)) , (du/dx)=((−4)/((x−1)^2 )) x=((u+3)/(u−1)) ⇒−(1/4)∫(du/(((4/(u−1)))(((4u)/(u−1)))^7 )) =−(1/4^9 )∫(((u−1)^8 )/u^7 )du =∫(u−8+((28)/u)−((56)/u^2 )+((70)/u^3 )−((56)/u^4 )+((28)/u^5 )−(8/u^6 )+(1/u^7 ))du ×(−(1/4^9 )) =((u^2 /2)−8u+28ln∣u∣+((56)/u)−((35)/u^2 )+((56)/(3u^3 ))−(7/u^4 ) +(8/(5u^5 ))−(1/(6u^(6 ) )))×(−(1/(262144)))+c plug u=((x+3)/(x−1)) in ⇒∫(dx/((x−1)^3 (x+3)^7 )) =−(1/(262144))[(((((x+3)/(x−1)))^2 )/2)−((8(x+3))/(x−1))+28ln∣((x+3)/(x−1))∣ +((56(x−1))/(x+3))−35(((x−1)/(x+3)))^2 +((56)/3)(((x−1)/(x+3)))^3 − 7(((x−1)/(x+3)))^4 +(8/5)(((x−1)/(x+3)))^5 −(1/6)(((x−1)/(x+3)))^6 ]+c
∫dx(x−1)3(x+3)7letu=x+3x−1,dudx=−4(x−1)2x=u+3u−1⇒−14∫du(4u−1)(4uu−1)7=−149∫(u−1)8u7du=∫(u−8+28u−56u2+70u3−56u4+28u5−8u6+1u7)du×(−149)=(u22−8u+28ln∣u∣+56u−35u2+563u3−7u4+85u5−16u6)×(−1262144)+cplugu=x+3x−1in⇒∫dx(x−1)3(x+3)7=−1262144[(x+3x−1)22−8(x+3)x−1+28ln∣x+3x−1∣+56(x−1)x+3−35(x−1x+3)2+563(x−1x+3)3−7(x−1x+3)4+85(x−1x+3)5−16(x−1x+3)6]+c
Commented by Tony Lin last updated on 12/Feb/20
∫(dx/((x+a)^m (x+b)^n )) let u=((x+b)/(x+a)) , (du/dx)=((a−b)/((x+a)^2 )) x=((b−au)/(u−1)) ⇒(1/(a−b))∫(du/((((b−a)/(u−1)))^(m−2) [(((b−a)u)/(u−1))]^n )) =−(1/((b−a)^(m+n−1) ))∫ (((u−1)^(m+n−2) )/u^n )du =∫((Σ_(r=0) ^(m+n−2) C_r ^(m+n−2) (((x+b)/(x+a)))^(r−n) (−1)^(m+n−2−r) )/(−(b−a)^(m+n−1) ))d(((x+b)/(x+a)))
∫dx(x+a)m(x+b)nletu=x+bx+a,dudx=a−b(x+a)2x=b−auu−1⇒1a−b∫du(b−au−1)m−2[(b−a)uu−1]n=−1(b−a)m+n−1∫(u−1)m+n−2undu=∫∑m+n−2r=0Crm+n−2(x+bx+a)r−n(−1)m+n−2−r−(b−a)m+n−1d(x+bx+a)
Commented by mathmax by abdo last updated on 12/Feb/20
thank you sir.
thankyousir.
Commented by MJS last updated on 12/Feb/20
on my path the hard work is solving the system for the coefficients c_j on your path it is to compose all those fractions if needed
onmypaththehardworkissolvingthesystemforthecoefficientscjonyourpathitistocomposeallthosefractionsifneeded
Commented by mathmax by abdo last updated on 12/Feb/20
∫F(x)dx =∫ (dx/((x−1)^3 (x+3)^7 )) =∫ (dx/((((x−1)/(x+3)))^3 (x+3)^(10) )) changement ((x−1)/(x+3))=t give x−1=tx+3t ⇒(1−t)x=3t+1 ⇒x=((3t+1)/(1−t)) ⇒ x+3=3+((3t+1)/(1−t)) =((3−3t+3t+1)/(1−t)) =(4/(1−t)) ⇒dx =(4/((1−t)^2 )) ⇒ ∫F(x)dx =∫ ((4dt)/((t−1)^2 t^3 ((4/(1−t)))^(10) )) =4 ∫ (((t−1)^(10) )/(4^(10) (t−1)^2 t^3 ))dt =(1/4^9 ) ∫ (((t−1)^8 )/t^3 )dt =(1/4^9 )∫ ((Σ_(k=0) ^8 C_8 ^k t^k (−1)^(8−k) )/t^3 )dt =(1/4^9 ) ∫ ((1−C_8 ^1 t +C_8 ^2 t^2 −C_8 ^3 t^3 +C_8 ^4 t^4 −C_8 ^5 t^5 +C_8 ^6 t^6 −C_8 ^7 t^7 +C_8 ^8 t^8 )/t^3 )dt =(1/4^9 ) ∫ (dt/t^3 ) −(C_8 ^1 /4^9 ) ∫ (dt/t^2 ) +(C_8 ^2 /4^9 )∫ (dt/t) −(C_8 ^3 /4^9 )∫dt (C_8 ^4 /4^9 ) ∫ tdt−(C_8 ^5 /4^9 ) ∫ t^2 dt +(C_8 ^6 /4^9 )∫ t^3 dt −(C_8 ^7 /4^9 ) ∫ t^4 dt +(C_8 ^8 /4^9 )∫ t^5 dt =−(1/(2×4^9 ×t^2 )) +(8/(4^9 ×t)) +(C_8 ^2 /4^9 )ln∣t∣−(C_8 ^3 /4^9 )t +(C_8 ^4 /(2.4^9 ))t^2 −(C_8 ^5 /(3.4^9 ))t^3 +(C_8 ^6 /4^(10) )t^4 −(C_8 ^7 /(5.4^9 ))t^5 +(1/(6.4^9 ))t^6 +C =−(1/(2.4^9 ))(((x+3)/(x−1)))^2 +(8/4^9 )(((x+3)/(x−1)))+(C_8 ^2 /4^9 )ln∣((x−1)/(x+3))∣−(C_8 ^3 /4^9 )(((x−1)/(x+3))) +(C_8 ^4 /(2.4^9 ))(((x−1)/(x+3)))^2 −(C_8 ^5 /(3.4^9 ))(((x−1)/(x+3)))^3 +(C_8 ^6 /4^(10) )(((x−1)/(x+3)))^4 −(C_8 ^7 /(5.4^9 ))(((x−1)/(x+3)))^5 +(1/(6.4^9 ))(((x−1)/(x+3)))^6 +C
∫F(x)dx=∫dx(x−1)3(x+3)7=∫dx(x−1x+3)3(x+3)10changementx−1x+3=tgivex−1=tx+3t⇒(1−t)x=3t+1⇒x=3t+11−t⇒x+3=3+3t+11−t=3−3t+3t+11−t=41−t⇒dx=4(1−t)2⇒∫F(x)dx=∫4dt(t−1)2t3(41−t)10=4∫(t−1)10410(t−1)2t3dt=149∫(t−1)8t3dt=149∫∑k=08C8ktk(−1)8−kt3dt=149∫1−C81t+C82t2−C83t3+C84t4−C85t5+C86t6−C87t7+C88t8t3dt=149∫dtt3−C8149∫dtt2+C8249∫dtt−C8349∫dtC8449∫tdt−C8549∫t2dt+C8649∫t3dt−C8749∫t4dt+C8849∫t5dt=−12×49×t2+849×t+C8249ln∣t∣−C8349t+C842.49t2−C853.49t3+C86410t4−C875.49t5+16.49t6+C=−12.49(x+3x−1)2+849(x+3x−1)+C8249ln∣x−1x+3∣−C8349(x−1x+3)+C842.49(x−1x+3)2−C853.49(x−1x+3)3+C86410(x−1x+3)4−C875.49(x−1x+3)5+16.49(x−1x+3)6+C
Commented by mathmax by abdo last updated on 12/Feb/20
∫_2 ^(+∞) F(x)dx =−(1/(2.4^9 ))[(((x+3)/(x−1)))^2 ]_2 ^(+∞) +(8/4^9 )[(((x+3)/(x−1)))]_2 ^(+∞) +(C_8 ^2 /4^9 )[ln∣((x−1)/(x+3))∣]_2 ^(+∞) −(C_8 ^3 /4^9 )[(((x−1)/(x+3)))]_2 ^(+∞) +(C_8 ^4 /(2.4^9 ))[(((x−1)/(x+3)))^2 ]_2 ^(+∞) −(C_8 ^5 /(3.4^9 ))[(((x−1)/(x+3)))^3 ]_2 ^(+∞) +(C_8 ^6 /4^(10) )[(((x−1)/(x+3)))^4 ]_2 ^(+∞) −(C_8 ^7 /(5.4^9 ))[(((x−1)/(x+3)))^5 ]_2 ^(+∞) +(1/(6.4^9 ))[(((x−1)/(x+3)))^6 ]_2 ^(+∞) =−(1/(2.4^9 )){1−5^2 }+(8/4^9 ){1−5}+(C_8 ^2 /4^9 ){−ln((1/5))} −(C_8 ^3 /4^9 ){1−(1/5)}+(C_8 ^4 /(2.4^9 )){1−((1/5))^2 }−(C_8 ^5 /(3.4^9 )){1−((1/5))^3 } +(C_8 ^6 /4^(10) ){1−((1/5))^4 }−(C_8 ^7 /(5.4^9 )){1−((1/5))^5 } +(1/(6.4^9 )){1−((1/5))^6 }
∫2+∞F(x)dx=−12.49[(x+3x−1)2]2+∞+849[(x+3x−1)]2+∞+C8249[ln∣x−1x+3∣]2+∞−C8349[(x−1x+3)]2+∞+C842.49[(x−1x+3)2]2+∞−C853.49[(x−1x+3)3]2+∞+C86410[(x−1x+3)4]2+∞−C875.49[(x−1x+3)5]2+∞+16.49[(x−1x+3)6]2+∞=−12.49{1−52}+849{1−5}+C8249{−ln(15)}−C8349{1−15}+C842.49{1−(15)2}−C853.49{1−(15)3}+C86410{1−(15)4}−C875.49{1−(15)5}+16.49{1−(15)6}
Commented by mathmax by abdo last updated on 12/Feb/20
the avantage of this method is finding integral without passing by decomposition..!
theavantageofthismethodisfindingintegralwithoutpassingbydecomposition..!
Answered by MJS last updated on 12/Feb/20
Ostrogradski′s Method again (it′s faster in my opinion) ∫(dx/((x+3)^7 (x−1)^3 )) ∫((P(x))/(Q(x)))dx=((P_1 (x))/(Q_1 (x)))+∫((P_2 (x))/(Q_2 (x)))dx Q_1 (x)=gcd (Q(x), Q^′ (x))=(x+3)^6 (x−1)^2 oo Q_2 (x)=((Q(x))/(Q_1 (x)))=(x+3)(x−1) ((P(x))/(Q(x)))=(d/dx)[((P_1 (x))/(Q_1 (x)))]+((P_2 (x))/(Q_2 (x))) (1/((x+3)^7 (x−1)^3 ))=(d/dx)[((Σ_(j=0) ^7 c_j x^j )/((x+3)^6 (x−1)^2 ))]+((d_1 x+d_0 )/((x+3)(x−1))) ⇒ P_1 (x)=((105x^7 +1575x^6 +9065x^5 +23135x^4 +15491x^3 −38339x^2 −56405x+14653)/(245760)) P_2 (x)=(7/(16384)) ⇒ we only have to solve (7/(16384))∫(dx/((x+3)(x−1)))= =(7/(65536))ln ∣((x−1)/(x+3))∣ ∫(dx/((x+3)^7 (x−1)^3 ))= =((105x^7 +1575x^6 +9065x^5 +23135x^4 +15491x^3 −38339x^2 −56405x+14653)/(245760(x+3)^6 (x−1)^2 ))+(7/(65536))ln ∣((x−1)/(x+3))∣ +C ∫_2 ^∞ (dx/((x+3)^7 (x−1)^3 ))=((−517516+328125ln 5)/(3072000000))
Ostrogradski′sMethodagain(it′sfasterinmyopinion)∫dx(x+3)7(x−1)3∫P(x)Q(x)dx=P1(x)Q1(x)+∫P2(x)Q2(x)dxQ1(x)=gcd(Q(x),Q′(x))=(x+3)6(x−1)2ooQ2(x)=Q(x)Q1(x)=(x+3)(x−1)P(x)Q(x)=ddx[P1(x)Q1(x)]+P2(x)Q2(x)1(x+3)7(x−1)3=ddx[∑7j=0cjxj(x+3)6(x−1)2]+d1x+d0(x+3)(x−1)⇒P1(x)=105x7+1575x6+9065x5+23135x4+15491x3−38339x2−56405x+14653245760P2(x)=716384⇒weonlyhavetosolve716384∫dx(x+3)(x−1)==765536ln∣x−1x+3∣∫dx(x+3)7(x−1)3==105x7+1575x6+9065x5+23135x4+15491x3−38339x2−56405x+14653245760(x+3)6(x−1)2+765536ln∣x−1x+3∣+C∫∞2dx(x+3)7(x−1)3=−517516+328125ln53072000000
Commented by john santu last updated on 12/Feb/20
ostrogradski′s method can use any type integral?
ostrogradski′smethodcanuseanytypeintegral?
Commented by msup trace by abdo last updated on 12/Feb/20
thank you sir .
thankyousir.
Commented by MJS last updated on 12/Feb/20
no. P(x) and Q(x) must be polynomes. the degree of P(x) must be smaller than the degree of Q(x) and all coefficients must be real the degree of P_1 (x) will then be smaller than the degree of Q_1 (x), same for P_2 (x) and Q_2 (x) I think you can find information on wikipedia
no.P(x)andQ(x)mustbepolynomes.thedegreeofP(x)mustbesmallerthanthedegreeofQ(x)andallcoefficientsmustberealthedegreeofP1(x)willthenbesmallerthanthedegreeofQ1(x),sameforP2(x)andQ2(x)Ithinkyoucanfindinformationonwikipedia
Commented by jagoll last updated on 12/Feb/20
thank you mister
thankyoumister

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