# How do you find int x^3/(x^4-x^2-1)dx using partial fractions?

Oct 31, 2015

$\int \frac{{x}^{3}}{{x}^{4} - {x}^{2} - 1} \mathrm{dx} = \frac{5 + \sqrt{5}}{20} \ln | {x}^{2} - \frac{1 + \sqrt{5}}{2} | + \frac{5 - \sqrt{5}}{20} \ln \left({x}^{2} + \frac{\sqrt{5} - 1}{2}\right) + c ,$
where $c$ is the constant of integration.

#### Explanation:

To get the partial fraction, we need to factorize the denominator, which is ${x}^{4} - {x}^{2} - 1$. To do so, first observe that there are no terms with degree 1 or 3. Hence, we can make a substitution of $y = {x}^{2}$. So,

${x}^{4} - {x}^{2} - 1 = {y}^{2} - y - 1$,

which becomes a quadratic expression. Using either the quadratic formula or by completing the square, the roots to the equation ${y}^{2} - y - 1 = 0$ are $y = \frac{1 + \sqrt{5}}{2}$ and $y = \frac{1 - \sqrt{5}}{2}$.

Therefore,

${x}^{4} - {x}^{2} - 1 = {y}^{2} - y - 1$
$= \left(y - \frac{1 + \sqrt{5}}{2}\right) \left(y - \frac{1 - \sqrt{5}}{2}\right)$
$= \left({x}^{2} - \frac{1 + \sqrt{5}}{2}\right) \left({x}^{2} - \frac{1 - \sqrt{5}}{2}\right)$
$= \left(x + \sqrt{\frac{1 + \sqrt{5}}{2}}\right) \left(x - \sqrt{\frac{1 + \sqrt{5}}{2}}\right) \left({x}^{2} + \frac{\sqrt{5} - 1}{2}\right)$.

Now that the denominator has been completely factorized, we can proceed with the partial fractions. We first note that the numerator is of a smaller degree than the denominator. We can thus write
$\frac{{x}^{3}}{{x}^{4} - {x}^{2} - 1} \equiv \frac{A}{x + \sqrt{\frac{1 + \sqrt{5}}{2}}}$
$+ \frac{B}{x - \sqrt{\frac{1 + \sqrt{5}}{2}}} + \frac{C x + D}{{x}^{2} + \frac{\sqrt{5} - 1}{2}}$,
where $A$, $B$, $C$ and $D$ are real constants to be determined.

By multiplying both sides with $\left({x}^{4} - {x}^{2} - 1\right)$, we get
${x}^{3} \equiv A \left(x - \sqrt{\frac{1 + \sqrt{5}}{2}}\right) \left({x}^{2} + \frac{\sqrt{5} - 1}{2}\right)$
$+ B \left(x + \sqrt{\frac{1 + \sqrt{5}}{2}}\right) \left({x}^{2} + \frac{\sqrt{5} - 1}{2}\right)$
$+ \left(C x + D\right) \left({x}^{2} - \frac{1 + \sqrt{5}}{2}\right)$.

To find $A$, substitute a value of $x$ that will cancel out the terms with $B$, $C$ and $D$. In this case, substitute $x = - \sqrt{\frac{1 + \sqrt{5}}{2}}$.
(-sqrt{frac{1+sqrt{5}}{2}})^3=A((-sqrt{frac{1+sqrt{5}}{2}})
-sqrt{frac{1+sqrt{5}}{2}})((-sqrt{frac{1+sqrt{5}}{2}})^2+frac{sqrt{5}-1}{2})

We get $A = \frac{5 + \sqrt{5}}{20}$.

Similarly, to find $B$, substitute a value of $x$ that will cancel out the terms with $A$, $C$ and $D$. In this case, substitute $x = \sqrt{\frac{1 + \sqrt{5}}{2}}$.
(sqrt{frac{1+sqrt{5}}{2}})^3=B((sqrt{frac{1+sqrt{5}}{2}})
+sqrt{frac{1+sqrt{5}}{2}})((sqrt{frac{1+sqrt{5}}{2}})^2+frac{sqrt{5}-1}{2})

We get $B = \frac{5 + \sqrt{5}}{20}$.

Now to find $C$ and $D$, the approach is slightly different. To find $D$, we need to find a real number $x$ to substitute such that the terms with $A$, $B$ and $C$ will disappear. However, there are no such $x$. But not to worry, since we have already found $A$ and $B$, a value of $x$ that will eliminate the terms with $C$ only will suffice. In this case, substitute $x = 0$ .
${\left(0\right)}^{3} = \frac{5 + \sqrt{5}}{20} \left(\left(0\right) - \sqrt{\frac{1 + \sqrt{5}}{2}}\right) \left({\left(0\right)}^{2} + \frac{\sqrt{5} - 1}{2}\right)$
$+ \frac{5 + \sqrt{5}}{20} \left(\left(0\right) + \sqrt{\frac{1 + \sqrt{5}}{2}}\right) \left({\left(0\right)}^{2} + \frac{\sqrt{5} - 1}{2}\right)$
$+ \left(C \left(0\right) + D\right) \left({\left(0\right)}^{2} - \frac{1 + \sqrt{5}}{2}\right)$

We get $D = 0$.

To find $C$, we can compare the coefficients of the ${x}^{3}$ term.

$1 = A + B + C$

We get $C = \frac{5 - \sqrt{5}}{10}$.

Hence,
$\frac{{x}^{3}}{{x}^{4} - {x}^{2} - 1} \equiv \frac{\frac{5 + \sqrt{5}}{20}}{x + \sqrt{\frac{1 + \sqrt{5}}{2}}}$
$+ \frac{\frac{5 + \sqrt{5}}{20}}{x - \sqrt{\frac{1 + \sqrt{5}}{2}}} + \frac{\left(\frac{5 - \sqrt{5}}{10}\right) x}{{x}^{2} + \frac{\sqrt{5} - 1}{2}}$.

Now, we proceed with the integration.
$\int \frac{{x}^{3}}{{x}^{4} - {x}^{2} - 1} \mathrm{dx} = \int \frac{\frac{5 + \sqrt{5}}{20}}{x + \sqrt{\frac{1 + \sqrt{5}}{2}}} \mathrm{dx}$
$+ \int \frac{\frac{5 + \sqrt{5}}{20}}{x - \sqrt{\frac{1 + \sqrt{5}}{2}}} \mathrm{dx} + \int \frac{\left(\frac{5 - \sqrt{5}}{10}\right) x}{{x}^{2} + \frac{\sqrt{5} - 1}{2}} \mathrm{dx}$

$= \frac{5 + \sqrt{5}}{20} \ln | x + \sqrt{\frac{1 + \sqrt{5}}{2}} | + \frac{5 + \sqrt{5}}{20} \ln | x - \sqrt{\frac{1 + \sqrt{5}}{2}} |$
$+ \int \frac{\left(\frac{5 - \sqrt{5}}{10}\right) x}{{x}^{2} + \frac{\sqrt{5} - 1}{2}} \mathrm{dx}$

$= \frac{5 + \sqrt{5}}{20} \ln | {x}^{2} - \frac{1 + \sqrt{5}}{2} | + \int \frac{\left(\frac{5 - \sqrt{5}}{10}\right) x}{{x}^{2} + \frac{\sqrt{5} - 1}{2}} \mathrm{dx}$

$= \frac{5 + \sqrt{5}}{20} \ln | {x}^{2} - \frac{1 + \sqrt{5}}{2} | + \left(\frac{5 - \sqrt{5}}{20}\right) \int \frac{2 x}{{x}^{2} + \frac{\sqrt{5} - 1}{2}} \mathrm{dx}$

$= \frac{5 + \sqrt{5}}{20} \ln | {x}^{2} - \frac{1 + \sqrt{5}}{2} | + \frac{5 - \sqrt{5}}{20} \ln \left({x}^{2} + \frac{\sqrt{5} - 1}{2}\right)$
$+ c$, where $c$ is the constant of integration.