How do you integrate $\int \frac{1}{(x^{3}) - 1}$ $int 1/[(x^3)-1]$ using partial fractions?

Question

How do you integrate $\int \frac{1}{(x^{3}) - 1}$ $int 1/[(x^3)-1]$ using partial fractions?

1 Answer

Impact of this question

3287 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-31T17:39:42

The integral is $\frac{1}{3} ln (x - 1) - \frac{1}{6} ln (x^{2} + x + 1) - \frac{1}{\sqrt{3}} arctan (\frac{2 x + 1}{\sqrt{3}})$ $1/3ln(x-1)-1/6ln(x^2+x+1)-1/(sqrt3)arctan((2x+1)/sqrt3)$

Explanation:

Let's factorise the denominator
$x^{3} - 1 = (x - 1) (x^{2} + x + 1)$ $x^3-1=(x-1)(x^2+x+1)$
and the decomposition in partial fractions
$\frac{1}{x^{3} - 1} = \frac{A}{x - 1} + \frac{B x + C}{x^{2} + x + 1}$ $1/(x^3-1)=A/(x-1)+(Bx+C)/(x^2+x+1)$
$1 = A (x^{2} + x + 1) + (B x + C) (x - 1)$ $1=A(x^2+x+1)+ (Bx+C)(x-1)$
If $x = 0$ $x=0$ $\Rightarrow$ $=>$ $1 = A - C$ $1=A-C$
coefficients of x^2, $0 = A + B$ $0=A+B$
Coefficients of x, $0 = A - B + C$ $0=A-B+C$
Solving, we found $A = \frac{1}{3}, B = - \frac{1}{3}, C = - \frac{2}{3}$ $A=1/3, B=-1/3, C=-2/3$
so $\int \frac{d x}{x^{3} - 1} = \frac{1}{3} \int \frac{d x}{x - 1} - \frac{1}{3} \int \frac{(x + 2) d x}{x^{2} + x + 1}$ $intdx/(x^3-1)=1/3intdx/(x-1)-1/3int((x+2)dx)/(x^2+x+1)$
$\int \frac{d x}{x - 1} = ln (x - 1)$ $intdx/(x-1)=ln(x-1)$
$\int \frac{(x + 2) d x}{x^{2} + x + 1} = \frac{1}{2} \int \frac{(2 x + 1) d x}{x^{2} + x + 1} + \frac{3}{2} \int \frac{d x}{x^{2} + x + 1}$ $int((x+2)dx)/(x^2+x+1)=1/2int((2x+1)dx)/(x^2+x+1)+3/2intdx/(x^2+x+1)$
$\frac{1}{2} \int \frac{(2 x + 1) d x}{x^{2} + x + 1} = \frac{1}{2} ln (x^{2} + x + 1)$ $1/2int((2x+1)dx)/(x^2+x+1)=1/2ln(x^2+x+1)$
$\int \frac{d x}{x^{2} + x + 1} = \int \frac{d x}{{(x + \frac{1}{2})}^{2} + \frac{3}{4}}$ $intdx/(x^2+x+1)=intdx/((x+1/2)^2+3/4)$
let's do the substitution, $u = \frac{2 x + 1}{\sqrt{3}}$ $u=(2x+1)/sqrt3$ $\Rightarrow$ $=>$ $\frac{d u}{d x} = \frac{2}{\sqrt{3}}$ $(du)/dx=2/sqrt3$
$\int \frac{d x}{{(x + \frac{1}{2})}^{2} + \frac{3}{4}} = \frac{2}{\sqrt{3}} \int \frac{d u}{u^{2} + 1} = \frac{2}{\sqrt{3}} arctan u$ $intdx/((x+1/2)^2+3/4)=2/sqrt3int(du)/(u^2+1)=2/sqrt3arctanu$

Science

Math

Humanities

... and beyond

How do you integrate $\int \frac{1}{(x^{3}) - 1}$ $int 1/[(x^3)-1]$ using partial fractions?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

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

1 Answer

Explanation:

Related questions

Impact of this question

How do you integrate $\int \frac{1}{(x^{3}) - 1}$ $int 1/[(x^3)-1]$ using partial fractions?