How do you integrate int (2x)/(1-x^3)∫2x1−x3 using partial fractions?

2 Answers
Shwetank Mauria
Jul 20, 2017

int(2x)/(1-x^3)dx=-2/3ln|x-1|+1/3ln|x^2+x+1|+2/sqrt3tan^(-1)((2x+1)/sqrt3)+c∫2x1−x3dx=−23ln|x−1|+13ln∣∣x2+x+1∣∣+2√3tan−1(2x+1√3)+c

Explanation:

As 1-x^3=(1-x)(1+x+x^2)1−x3=(1−x)(1+x+x2) let

(2x)/(1-x^3)=A/(1-x)+(Bx+C)/(1+x+x^2)2x1−x3=A1−x+Bx+C1+x+x2

= (A+Ax+Ax^2-Bx^2+Bx-Cx+C)/(1-x^3)A+Ax+Ax2−Bx2+Bx−Cx+C1−x3

= ((A-B)x^2+(A+B-C)x+(A+C))/(1-x^3)(A−B)x2+(A+B−C)x+(A+C)1−x3

Hence A-B=0A−B=0 i.e. B=AB=A, A+C=0A+C=0 i.e. C=-AC=−A

and A+B-C=2A+B−C=2 i.e. A+A+A=2A+A+A=2 i.e. A=2/3A=23

and B=2/3B=23 and C=-2/3C=−23

and int(2x)/(1-x^3)dx=-2/3int(dx)/(x-1)+2/3int(x-1)/(x^2+x+1)∫2x1−x3dx=−23∫dxx−1+23∫x−1x2+x+1

= -2/3int(dx)/(x-1)+1/3int(2x+1)/(x^2+x+1)dx+int1/(x^2+x+1)dx−23∫dxx−1+13∫2x+1x2+x+1dx+∫1x2+x+1dx

= -2/3ln|x-1|+1/3ln|x^2+x+1|+int1/(x^2+x+1)dx−23ln|x−1|+13ln∣∣x2+x+1∣∣+∫1x2+x+1dx

Now int1/(x^2+x+1)dx=int1/((x+1/2)^2+3/4)dx∫1x2+x+1dx=∫1(x+12)2+34dx

and putting u=x+1/2u=x+12 and du=dxdu=dx above becomes

int1/(u^2+3/4)du∫1u2+34du and as int1/(x^2+a^2)dx=1/atan^(-1)(x/a)∫1x2+a2dx=1atan−1(xa)

Hence int1/(u^2+3/4)du=1/(sqrt3/2)tan^(-1)(u/(sqrt3/2))∫1u2+34du=1√32tan−1⎛⎜⎝u√32⎞⎟⎠

and int1/(x^2+x+1)dx=2/sqrt3tan^(-1)((2x+1)/sqrt3)∫1x2+x+1dx=2√3tan−1(2x+1√3)

and int(2x)/(1-x^3)dx=-2/3ln|x-1|+1/3ln|x^2+x+1|+2/sqrt3tan^(-1)((2x+1)/sqrt3)+c∫2x1−x3dx=−23ln|x−1|+13ln∣∣x2+x+1∣∣+2√3tan−1(2x+1√3)+c

Andrea S.
Jul 20, 2017

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∫2x1−x3dx=−23ln|1−x|+13ln∣∣1+x+x2∣∣−2√3arctan(2x+1√3)+C

Explanation:

Factorize the denominator:

(1-x^3) = (1-x)(1+x+x^2)(1−x3)=(1−x)(1+x+x2)

Write the integrand as:

x/(1-x^3) = A/(1-x)+ (Bx+C)/(1+x+x^2)x1−x3=A1−x+Bx+C1+x+x2

x/(1-x^3) = (A(1+x+x^2)+(Bx+C)(1-x))/((1-x)(1+x+x^2))x1−x3=A(1+x+x2)+(Bx+C)(1−x)(1−x)(1+x+x2)

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

x = A+Ax+Ax^2+Bx+C-Bx^2-Cxx=A+Ax+Ax2+Bx+C−Bx2−Cx

x = (A-B)x^2 + (A+B-C)x+(A+C)x=(A−B)x2+(A+B−C)x+(A+C)

and equating the coefficients of the same degree in xx:

{(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

Related questions
See all questions in Integral by Partial Fractions
Impact of this question
1938 views around the world
You can reuse this answer
Creative Commons License