Question

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

Answer 1

`int(2x)/(1-x^3)dx=-2/3ln|x-1|+1/3ln|x^2+x+1|+2/sqrt3tan^(-1)((2x+1)/sqrt3)+c`

Explanation:

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`

Answer 2

`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`

Explanation:

Factorize the denominator:

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

Write the integrand as:

`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))`

As the denominators are equal, so must be the numerators:

`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`

Solve separately 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 solve:

`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 finally:

`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`

Putting it all together:

`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`