# How do you use partial fractions to find the integral int (x^2)/(x^4-2x^2-8)dx?

Jan 26, 2017

The integral equals $\frac{1}{6} \ln | x - 2 | - \frac{1}{6} \ln | x + 2 | + \arctan \frac{\frac{x}{\sqrt{2}}}{3 \sqrt{2}} + C$

#### Explanation:

Recall that ${x}^{4} - 2 {x}^{2} - 8$ can be factored as $\left({x}^{2} - 4\right) \left({x}^{2} + 2\right) = \left(x + 2\right) \left(x - 2\right) \left({x}^{2} + 2\right)$.

Hence:

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

$A \left(x - 2\right) \left({x}^{2} + 2\right) + B \left(x + 2\right) \left({x}^{2} + 2\right) + \left(C x + D\right) \left({x}^{2} - 4\right) = {x}^{2}$

$A \left({x}^{3} - 2 {x}^{2} + 2 x - 4\right) + B \left({x}^{3} + 2 {x}^{2} + 2 x + 4\right) + C {x}^{3} + D {x}^{2} - 4 C x - 4 D = {x}^{2}$

$A {x}^{3} - 2 A {x}^{2} + 2 A x - 4 A + B {x}^{3} + 2 B {x}^{2} + 2 B x + 4 B + C {x}^{3} + D {x}^{2} - 4 C x - 4 D = {x}^{2}$

Write a system of equations.

$\left\{\begin{matrix}A + B + C = 0 \\ 2 B - 2 A + D = 1 \\ 2 A + 2 B - 4 C = 0 \\ 4 B - 4 A - 4 D = 0\end{matrix}\right.$

After solving, either using a computer algebra system or tedious algebra, you should get that $A = - \frac{1}{6}$, $B = \frac{1}{6}$, $C = 0$ and $D = \frac{1}{3}$.

$= \int - \frac{1}{6 \left(x + 2\right)} + \frac{1}{6 \left(x - 2\right)} + \frac{1}{3 \left({x}^{2} + 2\right)} \mathrm{dx}$

The first two terms can be integrated using the rule $\int \frac{1}{x} \mathrm{dx} = \ln | x | + C$. Therefore, we are left with:

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

We need to recognize the last integral as one integrable by arctangent. Use the substitution $u = \frac{x}{\sqrt{2}}$.

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

The integral of $\frac{1}{{x}^{2} + 1} \mathrm{dx} = \arctan x$:

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

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

Put this together

$= \frac{1}{6} \ln | x - 2 | - \frac{1}{6} \ln | x + 2 | + \arctan \frac{\frac{x}{\sqrt{2}}}{3 \sqrt{2}} + C$

Hopefully this helps!