How do you integrate #int (2x)/(1-x^3)# using partial fractions?

Question

How do you integrate #int  (2x)/(1-x^3)# using partial fractions?

Isabella Ackerman · Accepted Answer

#int(2x)/(1-x^3)dx=-2/3ln|x-1|+1/3ln|x^2+x+1|+2/sqrt3tan^(-1)((2x+1)/sqrt3)+c# As #1-x^3=(1-x)(1+x+x^2)# let #(2x)/(1-x^3)=A/(1-x)+(Bx+C)/(1+x+x^2)# = #(A+Ax+Ax^2-Bx^2+Bx-Cx+C)/(1-x^3)# = #((A-B)x^2+(A+B-C)x+(A+C))/(1-x^3)# Hence #A-B=0# i.e. #B=A#, #A+C=0# i.e. #C=-A# and #A+B-C=2# i.e. #A+A+A=2# i.e. #A=2/3# and #B=2/3# and #C=-2/3# and #int(2x)/(1-x^3)dx=-2/3int(dx)/(x-1)+2/3int(x-1)/(x^2+x+1)# = #-2/3int(dx)/(x-1)+1/3int(2x+1)/(x^2+x+1)dx+int1/(x^2+x+1)dx# = #-2/3ln|x-1|+1/3ln|x^2+x+1|+int1/(x^2+x+1)dx# Now #int1/(x^2+x+1)dx=int1/((x+1/2)^2+3/4)dx# and putting #u=x+1/2# and #du=dx# above becomes #int1/(u^2+3/4)du# and as #int1/(x^2+a^2)dx=1/atan^(-1)(x/a)# Hence #int1/(u^2+3/4)du=1/(sqrt3/2)tan^(-1)(u/(sqrt3/2))# and #int1/(x^2+x+1)dx=2/sqrt3tan^(-1)((2x+1)/sqrt3)# and #int(2x)/(1-x^3)dx=-2/3ln|x-1|+1/3ln|x^2+x+1|+2/sqrt3tan^(-1)((2x+1)/sqrt3)+c#

Emily Ammerman · Answer

#int (2x)/(1-x^3)dx = -2/3 ln abs (1-x) +1/3 ln abs (1+x+x^2) -2/sqrt3arctan((2x+1)/sqrt3)+C#

Divide the denominator by two:

#(1-x^3) = (1-x)(1+x+x^2)#

For the integrand, write:

#x/(1-x^3) = A/(1-x)+ (Bx+C)/(1+x+x^2)#

#x/(1-x^3) = (A(1+x+x^2)+(Bx+C)(1-x))/((1-x)(1+x+x^2))#

The numerators must be equal since the denominators are:

#x = A+Ax+Ax^2+Bx+C-Bx^2-Cx#

#x = (A-B)x^2 + (A+B-C)x+(A+C)#

and equating the coefficients of the same degree in #x#:

#{(A-B=0),(A+B-C=1),(A+C=0):}#

#{(A=B),(2A-C=1),(C=-A):}#

#{(A=1/3),(B=1/3),(C=-1/3):}#

Then:

#x/(1-x^3) = 1/3 1/(1-x)+ 1/3 (x-1)/(1+x+x^2)#

and:

#int (2x)/(1-x^3)dx = 2/3int dx/(1-x) +2/3int (x-1)/(1+x+x^2)dx#

Separately solve the integrals:

#2/3int dx/(1-x) = - 2/3int (d(1-x))/(1-x) = - 2/3ln abs (1-x) + C_1#

Split the other noting that #d(1+x+x^2) = 1+2x#:

#2/3int (x-1)/(1+x+x^2)dx = 1/3 int (2x+1-3)/(1+x+x^2)dx#

#2/3int (x-1)/(1+x+x^2)dx = 1/3int (2x+1)/(1+x+x^2)dx - int 1/(1+x+x^2)dx #

Now resolve:

#1/3int (2x+1)/(1+x+x^2)dx = 1/3int (d(1+x+x^2))/(1+x+x^2) = 1/3ln abs (1+x+x^2) + C_2#

and lastly:

#int 1/(1+x+x^2)dx = int dx/(3/4+ (x+1/2)^2)#

#int 1/(1+x+x^2)dx = 4/3 int dx/(1+ ((2x+1)/sqrt3)^2)#

#int 1/(1+x+x^2)dx = 2/sqrt3 int (d((2x+1)/sqrt3))/(1+ ((2x+1)/sqrt3)^2)#

#int 1/(1+x+x^2)dx = 2/sqrt3 arctan((2x+1)/sqrt3)+C_3#

Combining everything:

#int (2x)/(1-x^3)dx = -2/3 ln abs (1-x) +1/3 ln abs (1+x+x^2) -2/sqrt3arctan((2x+1)/sqrt3)+C#

Emilia Aguilar · Answer

undefined To integrate ∫(2x)/(1-x^3) using partial fractions:

1. Factor the denominator: 1 - x^3 = (1 - x)(1 + x + x^2).

2. Write the fraction as a sum of partial fractions with undetermined coefficients:

(2x)/(1-x^3) = A/(1 - x) + (Bx + C)/(1 + x + x^2).

3. Find the values of A, B, and C by equating the numerators of the original expression and the partial fractions expression:

2x = A(1 + x + x^2) + (Bx + C)(1 - x).

4. Expand the right side and equate coefficients of like terms.

5. Solve the resulting system of equations for A, B, and C.

6. Once you have found A, B, and C, integrate each partial fraction separately.

7. Finally, add the integrated partial fractions together to get the result of the original integral.