Question Number 102922 by mohammad17 last updated on 11/Jul/20

∫ (dx/(x^3 +3x−5))  ?

Commented byprakash jain last updated on 11/Jul/20

The approach you take to solve  the question  factorize denominator  (x−α)(x−β)(x−γ)  partial fraction  (a/(x−α))+(b/(x−β))+(c/(x−γ))  integrate and simplify

Answered by 1549442205 last updated on 12/Jul/20

Puttimg x=y−(1/y)⇒x^3 =y^3 −3y+(3/y)−(1/y^3 )  x^3 +3x−5=0⇔y^3 −3y+(3/y)−(1/y^3 )+3y−(3/y)−5=0  ⇔y^3 −(1/y^3 )−5=0⇔y^6 −5y^3 −1=0.This  a quadratic eqs for y^3 .Δ=29  y^3 =((5±(√(29)))/2)⇔y= { ((y_1 = ^3 (√((5+(√(29)))/2)))),((y_2 = ^3 (√((5−(√(29)))/2)))) :}  x= ^3 (√((5+(√(29)))/2))−(1/( ^3 (√((5+(√(29)))/2))))  = ^3 (√((5+(√(29)))/2))+^3 (√((5−(√(29)))/2)) is unique  root.Hence x^3 +3x−5=(x−x_0 )(x^2 +ax+b)  ⇔x^3 +3x−5=x^3 +(a−x_0 )x^2 +(b−ax_0 )x+bx_0   ⇔ { ((bx_0 =−5)),((b−ax_0 =3)),((a−x_0 =0)) :}  Hence, a=x_0 = ^3 (√((5+(√(29)))/2))+ ^3 (√((5−(√(29)))/2)) ≈1.15(1)  b=((−5)/( ^3 (√((5+(√(29)))/2))+ ^3 (√((5−(√(29)))/2))))≈−4.33 (2)  we find A,B,C so that  (1/(x^3 +3x−5))=(A/(x−x_0 ))+((Bx+C)/(x^2 +ax+b))  ⇔(A+B)x^2 +(Aa−Bx_0 +C)x+Ab−Cx_0 =1   { ((A+B=0)),((Aa−Bx_0 +C=0)),((Ab−Cx_0 =1)) :} ⇔ { ((A(a+x_0 )+C=0)),((Ab−Cx_0 =1)) :}  ⇒Ab+A(a+x_0 )=1⇒A=(1/(a+b+x_0 ))  B=((−1)/(a+b+x_0 )),C=((b/(a+b+x_0 ))−1)/x_0   C=((−a−x_0 )/((a+b+x_0 )x_0 ))  F=∫(dx/(x^3 +3x−5))=∫(A/(x−x_0 ))dx+∫((Bx+C)/(x^2 +ax+b))dx  =Aln∣x−x_0 ∣+(B/2)∫((2x+a)/(x^2 +ax+b))dx+∫((C−((aB)/2))/(x^2 +ax+b))dx  =Aln∣x−x_0 ∣+(B/2)ln∣x^2 +ax+b∣+(C−((aB)/2))∫(dx/((x+(a/2))^2 −((a^2 /4)−b)))  =Aln∣x−x_0 ∣+(B/2)ln∣x^2 +ax+b∣+(C−((aB)/2))×(2/(√(a^2 −4b)))ln∣((x+(a/2)−((√(a^2 −4b))/2))/(x+(a/2)+((√(a^2 −4b))/2)))∣