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

Jul 20, 2017

$\int \frac{2 x}{1 - {x}^{3}} \mathrm{dx} = - \frac{2}{3} \ln | x - 1 | + \frac{1}{3} \ln | {x}^{2} + x + 1 | + \frac{2}{\sqrt{3}} {\tan}^{- 1} \left(\frac{2 x + 1}{\sqrt{3}}\right) + c$

Explanation:

As $1 - {x}^{3} = \left(1 - x\right) \left(1 + x + {x}^{2}\right)$ let

$\frac{2 x}{1 - {x}^{3}} = \frac{A}{1 - x} + \frac{B x + C}{1 + x + {x}^{2}}$

= $\frac{A + A x + A {x}^{2} - B {x}^{2} + B x - C x + C}{1 - {x}^{3}}$

= $\frac{\left(A - B\right) {x}^{2} + \left(A + B - C\right) x + \left(A + C\right)}{1 - {x}^{3}}$

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. $A = \frac{2}{3}$

and $B = \frac{2}{3}$ and $C = - \frac{2}{3}$

and $\int \frac{2 x}{1 - {x}^{3}} \mathrm{dx} = - \frac{2}{3} \int \frac{\mathrm{dx}}{x - 1} + \frac{2}{3} \int \frac{x - 1}{{x}^{2} + x + 1}$

= $- \frac{2}{3} \int \frac{\mathrm{dx}}{x - 1} + \frac{1}{3} \int \frac{2 x + 1}{{x}^{2} + x + 1} \mathrm{dx} + \int \frac{1}{{x}^{2} + x + 1} \mathrm{dx}$

= $- \frac{2}{3} \ln | x - 1 | + \frac{1}{3} \ln | {x}^{2} + x + 1 | + \int \frac{1}{{x}^{2} + x + 1} \mathrm{dx}$

Now $\int \frac{1}{{x}^{2} + x + 1} \mathrm{dx} = \int \frac{1}{{\left(x + \frac{1}{2}\right)}^{2} + \frac{3}{4}} \mathrm{dx}$

and putting $u = x + \frac{1}{2}$ and $\mathrm{du} = \mathrm{dx}$ above becomes

$\int \frac{1}{{u}^{2} + \frac{3}{4}} \mathrm{du}$ and as $\int \frac{1}{{x}^{2} + {a}^{2}} \mathrm{dx} = \frac{1}{a} {\tan}^{- 1} \left(\frac{x}{a}\right)$

Hence $\int \frac{1}{{u}^{2} + \frac{3}{4}} \mathrm{du} = \frac{1}{\frac{\sqrt{3}}{2}} {\tan}^{- 1} \left(\frac{u}{\frac{\sqrt{3}}{2}}\right)$

and $\int \frac{1}{{x}^{2} + x + 1} \mathrm{dx} = \frac{2}{\sqrt{3}} {\tan}^{- 1} \left(\frac{2 x + 1}{\sqrt{3}}\right)$

and $\int \frac{2 x}{1 - {x}^{3}} \mathrm{dx} = - \frac{2}{3} \ln | x - 1 | + \frac{1}{3} \ln | {x}^{2} + x + 1 | + \frac{2}{\sqrt{3}} {\tan}^{- 1} \left(\frac{2 x + 1}{\sqrt{3}}\right) + c$

Jul 20, 2017

$\int \frac{2 x}{1 - {x}^{3}} \mathrm{dx} = - \frac{2}{3} \ln \left\mid 1 - x \right\mid + \frac{1}{3} \ln \left\mid 1 + x + {x}^{2} \right\mid - \frac{2}{\sqrt{3}} \arctan \left(\frac{2 x + 1}{\sqrt{3}}\right) + C$

Explanation:

Factorize the denominator:

$\left(1 - {x}^{3}\right) = \left(1 - x\right) \left(1 + x + {x}^{2}\right)$

Write the integrand as:

$\frac{x}{1 - {x}^{3}} = \frac{A}{1 - x} + \frac{B x + C}{1 + x + {x}^{2}}$

$\frac{x}{1 - {x}^{3}} = \frac{A \left(1 + x + {x}^{2}\right) + \left(B x + C\right) \left(1 - x\right)}{\left(1 - x\right) \left(1 + x + {x}^{2}\right)}$

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

$x = A + A x + A {x}^{2} + B x + C - B {x}^{2} - C x$

$x = \left(A - B\right) {x}^{2} + \left(A + B - C\right) x + \left(A + C\right)$

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

$\left\{\begin{matrix}A - B = 0 \\ A + B - C = 1 \\ A + C = 0\end{matrix}\right.$

$\left\{\begin{matrix}A = B \\ 2 A - C = 1 \\ C = - A\end{matrix}\right.$

$\left\{\begin{matrix}A = \frac{1}{3} \\ B = \frac{1}{3} \\ C = - \frac{1}{3}\end{matrix}\right.$

Then:

$\frac{x}{1 - {x}^{3}} = \frac{1}{3} \frac{1}{1 - x} + \frac{1}{3} \frac{x - 1}{1 + x + {x}^{2}}$

and:

$\int \frac{2 x}{1 - {x}^{3}} \mathrm{dx} = \frac{2}{3} \int \frac{\mathrm{dx}}{1 - x} + \frac{2}{3} \int \frac{x - 1}{1 + x + {x}^{2}} \mathrm{dx}$

Solve separately the integrals:

$\frac{2}{3} \int \frac{\mathrm{dx}}{1 - x} = - \frac{2}{3} \int \frac{d \left(1 - x\right)}{1 - x} = - \frac{2}{3} \ln \left\mid 1 - x \right\mid + {C}_{1}$

Split the other noting that $d \left(1 + x + {x}^{2}\right) = 1 + 2 x$:

$\frac{2}{3} \int \frac{x - 1}{1 + x + {x}^{2}} \mathrm{dx} = \frac{1}{3} \int \frac{2 x + 1 - 3}{1 + x + {x}^{2}} \mathrm{dx}$

$\frac{2}{3} \int \frac{x - 1}{1 + x + {x}^{2}} \mathrm{dx} = \frac{1}{3} \int \frac{2 x + 1}{1 + x + {x}^{2}} \mathrm{dx} - \int \frac{1}{1 + x + {x}^{2}} \mathrm{dx}$

Now solve:

$\frac{1}{3} \int \frac{2 x + 1}{1 + x + {x}^{2}} \mathrm{dx} = \frac{1}{3} \int \frac{d \left(1 + x + {x}^{2}\right)}{1 + x + {x}^{2}} = \frac{1}{3} \ln \left\mid 1 + x + {x}^{2} \right\mid + {C}_{2}$

and finally:

$\int \frac{1}{1 + x + {x}^{2}} \mathrm{dx} = \int \frac{\mathrm{dx}}{\frac{3}{4} + {\left(x + \frac{1}{2}\right)}^{2}}$

$\int \frac{1}{1 + x + {x}^{2}} \mathrm{dx} = \frac{4}{3} \int \frac{\mathrm{dx}}{1 + {\left(\frac{2 x + 1}{\sqrt{3}}\right)}^{2}}$

$\int \frac{1}{1 + x + {x}^{2}} \mathrm{dx} = \frac{2}{\sqrt{3}} \int \frac{d \left(\frac{2 x + 1}{\sqrt{3}}\right)}{1 + {\left(\frac{2 x + 1}{\sqrt{3}}\right)}^{2}}$

$\int \frac{1}{1 + x + {x}^{2}} \mathrm{dx} = \frac{2}{\sqrt{3}} \arctan \left(\frac{2 x + 1}{\sqrt{3}}\right) + {C}_{3}$

Putting it all together:

$\int \frac{2 x}{1 - {x}^{3}} \mathrm{dx} = - \frac{2}{3} \ln \left\mid 1 - x \right\mid + \frac{1}{3} \ln \left\mid 1 + x + {x}^{2} \right\mid - \frac{2}{\sqrt{3}} \arctan \left(\frac{2 x + 1}{\sqrt{3}}\right) + C$