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Question Number 55129 by MJS last updated on 17/Feb/19

question 54995 reposted ∫((x^3 +x^2 ))^(1/3) dx=?

question54995reposted∫x3+x23dx=?

Commented by Meritguide1234 last updated on 18/Feb/19

Answered by MJS last updated on 18/Feb/19

∫((x^3 +x^2 ))^(1/3) dx= [t=(x)^(1/3) → dx=3t^2 dt] =3∫t^4 ((t^3 +1))^(1/3) dt= [u=(((t^3 +1))^(1/3) /t) → du=−t^2 (((t^3 +1)^2 ))^(1/3) dt] =−3∫(u^3 /((u−1)^3 (u^2 +u+1)^3 ))du= [Ostrogradski′s Method: ∫((P(u))/(Q(u)))du=((P_1 (u))/(Q_1 (u)))+∫((P_2 (u))/(Q_2 (u)))du Q_1 (u)=gcd(Q(u), Q′(u))=u^6 −2u^3 +1 Q_2 (u)=((Q(u))/(Q_1 (u)))=u^3 −1 to find P_1 (u) and P_2 (u) we differentiate both sides: ((P(u))/(Q(u)))=(d/du)[((P_1 (u))/(Q_1 (u)))]+((P_2 (u))/(Q_2 (u))) with P_1 (u)=au^5 +bu^4 +cu^3 +du^2 +eu+f and P_2 (u)=gu^2 +hu+i solving for these constants we get P_1 (u)=−(1/(18))u^4 −(1/9)u P_2 (u)=−(1/9)] =((u(u^3 +2))/(6(u−1)^2 (u^2 +u+1)^2 ))+(1/3)∫(du/((u−1)(u^2 +u+1))) the first part is solved and equals (1/6)(3x+1)((x^3 +x^2 ))^(1/3) the second part: (1/3)∫(du/((u−1)(u^2 +u+1)))=(1/9)∫(du/(u−1))−(1/9)∫((u+2)/(u^2 +u+1))du the first is simply = (1/9)ln (u−1) = =(1/9)ln ((((x+1))^(1/3) −(x)^(1/3) )/(x)^(1/3) ) the second part: −(1/9)∫((u+2)/(u^2 +u+1))du=−(1/(18))∫((2u+1)/(u^2 +u+1))du−(1/6)∫(du/(u^2 +u+1)) the first is simply = −(1/(18))ln (u^2 +u+1) = =−(1/(18))ln (1+(((x+1)/x))^(1/3) +(((((x+1)/x))^2 ))^(1/3) ) the second is simply = −((√3)/9)arctan (((2u+1)(√3))/3) = =−((√3)/9)arctan (((2((x+1))^(1/3) +(x)^(1/3) )(√3))/(3(x)^(1/3) )) so we get (1/6)(3x+1)((x^3 +x^2 ))^(1/3) +(1/9)ln ∣((((x+1))^(1/3) −(x)^(1/3) )/(x)^(1/3) )∣ −(1/(18))ln ∣1+(((x+1)/x))^(1/3) +(((((x+1)/x))^2 ))^(1/3) ∣ −((√3)/9)arctan (((2((x+1))^(1/3) +(x)^(1/3) )(√3))/(3(x)^(1/3) )) +C

∫x3+x23dx=[t=x3→dx=3t2dt]=3∫t4t3+13dt=[u=t3+13t→du=−t2(t3+1)23dt]=−3∫u3(u−1)3(u2+u+1)3du=[Ostrogradski′sMethod:∫P(u)Q(u)du=P1(u)Q1(u)+∫P2(u)Q2(u)duQ1(u)=gcd(Q(u),Q′(u))=u6−2u3+1Q2(u)=Q(u)Q1(u)=u3−1tofindP1(u)andP2(u)wedifferentiatebothsides:P(u)Q(u)=ddu[P1(u)Q1(u)]+P2(u)Q2(u)withP1(u)=au5+bu4+cu3+du2+eu+fandP2(u)=gu2+hu+isolvingfortheseconstantswegetP1(u)=−118u4−19uP2(u)=−19]=u(u3+2)6(u−1)2(u2+u+1)2+13∫du(u−1)(u2+u+1)thefirstpartissolvedandequals16(3x+1)x3+x23thesecondpart:13∫du(u−1)(u2+u+1)=19∫duu−1−19∫u+2u2+u+1duthefirstissimply=19ln(u−1)==19lnx+13−x3x3thesecondpart:−19∫u+2u2+u+1du=−118∫2u+1u2+u+1du−16∫duu2+u+1thefirstissimply=−118ln(u2+u+1)==−118ln(1+x+1x3+(x+1x)23)thesecondissimply=−39arctan(2u+1)33==−39arctan(2x+13+x3)33x3soweget16(3x+1)x3+x23+19ln∣x+13−x3x3∣−118ln∣1+x+1x3+(x+1x)23∣−39arctan(2x+13+x3)33x3+C

Commented by MJS last updated on 18/Feb/19

(1/9)ln ((((x+1))^(1/3) −(x)^(1/3) )/(x)^(1/3) ) =(1/9)ln (((x+1))^(1/3) −(x)^(1/3) ) −(1/(27))ln x −(1/(18))ln (1+(((x+1)/x))^(1/3) +(((((x+1)/x))^2 ))^(1/3) ) = =−(1/(18))ln (((x^2 )^(1/3) +((x^2 +x))^(1/3) +(((x+1)^2 ))^(1/3) )/(x^2 )^(1/3) ) = =−(1/(18))ln ((x^2 )^(1/3) +((x^2 +x))^(1/3) +(((x+1)^2 ))^(1/3) ) +(1/(27))ln x ∫((x^3 +x^2 ))^(1/3) dx=(1/6)(3x+1)((x^3 +x^2 ))^(1/3) +(1/9)ln (((x+1))^(1/3) −(x)^(1/3) ) −(1/(18))ln ((x^2 )^(1/3) +((x^2 +x))^(1/3) +(((x+1)^2 ))^(1/3) ) −((√3)/9)arctan (((2((x+1))^(1/3) +(x)^(1/3) )(√3))/(3(x)^(1/3) )) +C

19lnx+13−x3x3=19ln(x+13−x3)−127lnx−118ln(1+x+1x3+(x+1x)23)==−118lnx23+x2+x3+(x+1)23x23==−118ln(x23+x2+x3+(x+1)23)+127lnx∫x3+x23dx=16(3x+1)x3+x23+19ln(x+13−x3)−118ln(x23+x2+x3+(x+1)23)−39arctan(2x+13+x3)33x3+C

Commented by peter frank last updated on 18/Feb/19

thank you

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