# What is int (4-x ) / (x^3-6x +4 )?

##### 1 Answer
Mar 15, 2016

$\int \frac{4 - x}{{x}^{3} - 6 x + 4} \mathrm{dx} =$

$= \frac{1}{3} \ln \left\mid x - 2 \right\mid + \frac{- 1 - 2 \sqrt{3}}{6} \ln \left\mid x + 1 - \sqrt{3} \right\mid + \frac{- 1 + 2 \sqrt{3}}{6} \ln \left\mid x + 1 + \sqrt{3} \right\mid + C$

#### Explanation:

Express as a partial fraction decomposition first:

${x}^{3} - 6 x + 4$

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

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

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

Solve:

$\frac{4 - x}{{x}^{3} - 6 x + 4}$

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

$= \frac{A \left({x}^{2} + 2 x - 2\right) + B \left(x - 2\right) \left(x + 1 + \sqrt{3}\right) + C \left(x - 2\right) \left(x + 1 - \sqrt{3}\right)}{{x}^{3} - 6 x + 4}$

$= \frac{A \left({x}^{2} + 2 x - 2\right) + B \left({x}^{2} - \left(1 - \sqrt{3}\right) x - 2 \left(1 + \sqrt{3}\right)\right) + C \left({x}^{2} - \left(1 + \sqrt{3}\right) x - 2 \left(1 - \sqrt{3}\right)\right)}{{x}^{3} - 6 x + 4}$

$= \frac{\left(A + B + C\right) {x}^{2} + \left(2 A - \left(1 - \sqrt{3}\right) B - \left(1 + \sqrt{3}\right) C\right) x - 2 \left(A + \left(1 + \sqrt{3}\right) B + \left(1 - \sqrt{3}\right) C\right)}{{x}^{3} - 6 x + 4}$

Equating coefficients, we get the following simulataneous equations:

$\left(a\right) \textcolor{w h i t e}{-} A + B + C = 0$

$\left(b\right) \textcolor{w h i t e}{-} 2 A - \left(1 - \sqrt{3}\right) B - \left(1 + \sqrt{3}\right) C = - 1$

$\left(c\right) \textcolor{w h i t e}{-} A + \left(1 + \sqrt{3}\right) B + \left(1 - \sqrt{3}\right) C = - 2$

Subtracting $\left(c\right)$ from $\left(b\right)$, we get:

$\left(d\right) \textcolor{w h i t e}{-} A - 2 B - 2 C = 1$

Adding twice $\left(a\right)$ the first equation to $\left(d\right)$, we get:

$3 A = 1$

So $A = \frac{1}{3}$

Adding $\left(b\right) + \left(c\right)$ we get:

$\left(e\right) \textcolor{w h i t e}{-} 3 A + 2 \sqrt{3} B - 2 \sqrt{3} C = - 3$

We know $3 A = 1$, so this simplifies to:

$2 \sqrt{3} B - 2 \sqrt{3} C = - 4$

Hence:

$\left(f\right) \textcolor{w h i t e}{-} B - C = - \frac{2}{\sqrt{3}} = - 2 \frac{\sqrt{3}}{3}$

From $\left(a\right)$ we get:

$\left(g\right) \textcolor{w h i t e}{-} B + C = - \frac{1}{3}$

Then $\left(f\right) + \left(g\right)$ gives us:

$2 B = \frac{- 1 - 2 \sqrt{3}}{3}$

and $\left(g\right) - \left(f\right)$ gives us:

$2 C = \frac{- 1 + 2 \sqrt{3}}{3}$

Hence:

$\frac{4 - x}{{x}^{3} - 6 x + 4}$

$= \frac{1}{3 \left(x - 2\right)} + \frac{- 1 - 2 \sqrt{3}}{6 \left(x + 1 - \sqrt{3}\right)} + \frac{- 1 + 2 \sqrt{3}}{6 \left(x + 1 + \sqrt{3}\right)}$

Then use: $\int \frac{1}{t} \mathrm{dt} = \ln \left\mid t \right\mid + C$ to find:

$\int \frac{4 - x}{{x}^{3} - 6 x + 4} \mathrm{dx} =$

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

$= \frac{1}{3} \ln \left\mid x - 2 \right\mid + \frac{- 1 - 2 \sqrt{3}}{6} \ln \left\mid x + 1 - \sqrt{3} \right\mid + \frac{- 1 + 2 \sqrt{3}}{6} \ln \left\mid x + 1 + \sqrt{3} \right\mid + C$