# Evaluate the integral  int \ x/(a^3-x^3) \ dx?

Oct 18, 2017

$\int \setminus \frac{x}{{a}^{3} - {x}^{3}} \setminus \mathrm{dx} = \frac{1}{6 a} \ln | {x}^{2} + a x + {a}^{2} | - \frac{\sqrt{3}}{3 a} \setminus \arctan \left(\frac{2 x + a}{\sqrt{3} a}\right) - \frac{1}{3 a} \setminus \ln | x - a | + C$

#### Explanation:

We seek:

$I = \int \setminus \frac{x}{{a}^{3} - {x}^{3}} \setminus \mathrm{dx}$

We can factorize the integrand, and then decompose into partial fractions:

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

$\text{ } = \frac{A}{a - x} + \frac{B x + C}{{x}^{2} + a x + {a}^{2}}$

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

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

Where $A , B , C$ are constants that are to be determined. We can find them by substitutions (In practice we do this via the "cover up" method:

Put $x = a \implies a = A \left({a}^{2} + {a}^{2} + {a}^{2}\right) \implies A = \frac{1}{3 a}$
$C o e f f \left({x}^{2}\right) : 0 = A - B \implies B = \frac{1}{3 a}$
$C o e f f \left({x}^{0}\right) : 0 = A {a}^{2} + a C \implies C = - \frac{1}{3}$

So, we have:

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

So we can write the integral as:

$I = \frac{1}{3 a} \int \setminus \frac{x - a}{{x}^{2} + a x + {a}^{2}} - \frac{1}{x - a} \setminus \mathrm{dx}$
$\setminus \setminus = \frac{1}{3 a} \left\{\int \setminus \frac{x - a}{{x}^{2} + a x + {a}^{2}} \setminus \mathrm{dx} - \int \setminus \frac{1}{x - a} \setminus \mathrm{dx}\right\}$
$\setminus \setminus = \frac{1}{3 a} \int \setminus \frac{x - a}{{x}^{2} + a x + {a}^{2}} \setminus \mathrm{dx} - \frac{1}{3 a} \setminus \ln | x - a | + C$
$\setminus \setminus = \frac{1}{3 a} \setminus {I}_{2} - \frac{1}{3 a} \setminus \ln | x - a | + C$

Now, let us consider the remaining integral, ${I}_{2}$, which requires a little manipulation, as follows:

${I}_{2} = \int \setminus \frac{x - a}{{x}^{2} + a x + {a}^{2}} \setminus \mathrm{dx}$
$\setminus \setminus \setminus = \frac{1}{2} \setminus \int \setminus \frac{2 \left(x - a\right)}{{x}^{2} + a x + {a}^{2}} \setminus \mathrm{dx}$
$\setminus \setminus \setminus = \frac{1}{2} \setminus \int \setminus \frac{2 x - 2 a}{{x}^{2} + a x + {a}^{2}} \setminus \mathrm{dx}$
$\setminus \setminus \setminus = \frac{1}{2} \setminus \int \setminus \frac{2 x + a - 3 a}{{x}^{2} + a x + {a}^{2}} \setminus \mathrm{dx}$
$\setminus \setminus \setminus = \frac{1}{2} \setminus \int \setminus \frac{2 x + a}{{x}^{2} + a x + {a}^{2}} - \frac{3 a}{{x}^{2} + a x + {a}^{2}} \setminus \mathrm{dx}$
$\setminus \setminus \setminus = \frac{1}{2} \setminus \int \setminus \frac{2 x + a}{{x}^{2} + a x + {a}^{2}} - \frac{3 a}{2} \setminus \int \setminus \frac{1}{{x}^{2} + a x + {a}^{2}} \setminus \mathrm{dx}$
$\setminus \setminus \setminus = \frac{1}{2} \setminus \ln | {x}^{2} + a x + {a}^{2} | - \frac{3 a}{2} {I}_{3}$

Next, we consider, ${I}_{3}$, where

${I}_{3} = \int \setminus \frac{1}{{x}^{2} + a x + {a}^{2}} \setminus \mathrm{dx}$
$\setminus \setminus \setminus = \int \setminus \frac{1}{{\left(x + \frac{a}{2}\right)}^{2} - {\left(\frac{a}{2}\right)}^{2} + {a}^{2}} \setminus \mathrm{dx}$
$\setminus \setminus \setminus = \int \setminus \frac{1}{{\left(x + \frac{a}{2}\right)}^{2} - {a}^{2} / 4 + {a}^{2}} \setminus \mathrm{dx}$
$\setminus \setminus \setminus = \int \setminus \frac{1}{{\left(x + \frac{a}{2}\right)}^{2} + \frac{3 {a}^{2}}{4}} \setminus \mathrm{dx}$
$\setminus \setminus \setminus = \int \setminus \frac{1}{\frac{3 {a}^{2}}{4} \left(\left(\frac{4}{3 {a}^{2}}\right) {\left(x + \frac{a}{2}\right)}^{2} + 1\right)} \setminus \mathrm{dx}$
$\setminus \setminus \setminus = \frac{4}{3 {a}^{2}} \setminus \int \setminus \frac{1}{\left(\frac{4}{3 {a}^{2}}\right) {\left(x + \frac{a}{2}\right)}^{2} + 1} \setminus \mathrm{dx}$
$\setminus \setminus \setminus = \frac{4}{3 {a}^{2}} \setminus \int \setminus \frac{1}{{\left(\left(\frac{2}{\sqrt{3} a}\right) \left(x + \frac{a}{2}\right)\right)}^{2} + 1} \setminus \mathrm{dx}$
$\setminus \setminus \setminus = \frac{4}{3 {a}^{2}} \setminus \int \setminus \frac{1}{{\left(\frac{2}{\sqrt{3} a} x + \frac{1}{\sqrt{3}}\right)}^{2} + 1} \setminus \mathrm{dx}$

Now we can perform a substitution, Let

$u = \frac{2}{\sqrt{3} a} x + \frac{1}{\sqrt{3}} \implies \frac{\mathrm{du}}{\mathrm{dx}} = \frac{2}{\sqrt{3} a}$

Then substituting into ${I}_{3}$ we get:

${I}_{3} = \frac{4}{3 {a}^{2}} \setminus \int \setminus \frac{1}{{u}^{2} + 1} \setminus \left(\frac{\sqrt{3} a}{2}\right) \setminus \mathrm{du}$
$\setminus \setminus \setminus = \frac{2 \sqrt{3}}{3 a} \setminus \int \setminus \frac{1}{{u}^{2} + 1} \setminus \mathrm{du}$
$\setminus \setminus \setminus = \frac{2 \sqrt{3}}{3 a} \setminus \arctan \left(u\right)$
$\setminus \setminus \setminus = \frac{2 \sqrt{3}}{3 a} \setminus \arctan \left(\frac{2}{\sqrt{3} a} x + \frac{1}{\sqrt{3}}\right)$
$\setminus \setminus \setminus = \frac{2 \sqrt{3}}{3 a} \setminus \arctan \left(\frac{2 x + a}{\sqrt{3} a}\right)$

Combining this result with ${I}_{2}$ we get:

${I}_{2} = \frac{1}{2} \setminus \ln | {x}^{2} + a x + {a}^{2} | - \frac{3 a}{2} {I}_{3}$
$\setminus \setminus \setminus = \frac{1}{2} \setminus \ln | {x}^{2} + a x + {a}^{2} | - \frac{3 a}{2} \frac{2 \sqrt{3}}{3 a} \setminus \arctan \left(\frac{2 x + a}{\sqrt{3} a}\right)$
$\setminus \setminus \setminus = \frac{1}{2} \setminus \ln | {x}^{2} + a x + {a}^{2} | - \sqrt{3} \setminus \arctan \left(\frac{2 x + a}{\sqrt{3} a}\right)$

And finally combining this result with $I$ we get

$I = \frac{1}{3 a} \setminus {I}_{2} - \frac{1}{3 a} \setminus \ln | x - a | + C$
$\setminus \setminus = \frac{1}{3 a} \left\{\frac{1}{2} \setminus \ln | {x}^{2} + a x + {a}^{2} | - \sqrt{3} \setminus \arctan \left(\frac{2 x + a}{\sqrt{3} a}\right)\right\} - \frac{1}{3 a} \setminus \ln | x - a | + C$

$\setminus \setminus = \frac{1}{6 a} \ln | {x}^{2} + a x + {a}^{2} | - \frac{\sqrt{3}}{3 a} \setminus \arctan \left(\frac{2 x + a}{\sqrt{3} a}\right) - \frac{1}{3 a} \setminus \ln | x - a | + C$