# what is the integral ; { \int \frac{(x^2-x+6)}{(x^3-3x)}? }

## Question:

what is the integral ; {eq}\displaystyle \int \frac{(x^2-x+6)}{(x^3-3x)} \ dx {/eq}

## Integration By Partial Fraction:

Integration by a partial fraction is a method to solve the integral.

If we have to evaluate the integral of the form {eq}\displaystyle \int \frac{P(x)}{Q(x)}dx {/eq} where {eq}\displaystyle \frac{P(x)}{Q(x)} {/eq} is a proper rational fraction then first, we apply the partial fraction decomposition to split the fraction into a simpler form. After this integration can be carried out easily.

## Answer and Explanation:

**Given **

**{eq}\int \dfrac{(x^2-x+6)}{(x^3-3x)} {/eq} **

**We have to evaluate the integral. **

**{eq}\begin{align} \int \dfrac{(x^2-x+6)}{(x^3-3x)} dx &=\int \dfrac{(x^2-x+6)}{x(x^2-3)}\\ \\ &=\int \dfrac{(x^2-x+6)}{x(x-\sqrt{3})(x+\sqrt{3})} \end{align} {/eq} **

First, we are applying the partial fraction to split the fraction into a simpler form.

{eq}\begin{align} \dfrac{(x^2-x+6)}{x(x+\sqrt{3})(x-\sqrt{3})} &=\dfrac{A}{x}+\dfrac{B}{x+\sqrt{3}}+\dfrac{C}{x-\sqrt{3}}\\ \\ x^2-x+6 &=A\left(x+\sqrt{3}\right)\left(x-\sqrt{3}\right)+Bx\left(x-\sqrt{3}\right)+Cx\left(x+\sqrt{3}\right) \end{align} {/eq}

**Let** {eq}x=0 {/eq}

{eq}\begin{align} 0^2-0+6 &=A\left(0+\sqrt{3}\right)\left(0-\sqrt{3}\right)+B\cdot \:0\cdot \left(0-\sqrt{3}\right)+C\cdot \:0\cdot \left(0+\sqrt{3}\right)\\ \\ 6=-3A\\ \\ A=-2 \end{align} {/eq}

**Let** {eq}x+\sqrt{3}=0 \rightarrow x=-\sqrt{3} {/eq}

{eq}\begin{align} \left(-\sqrt{3}\right)^2-\left(-\sqrt{3}\right)+6 &=A\left(\left(-\sqrt{3}\right)+\sqrt{3}\right)\left(\left(-\sqrt{3}\right)-\sqrt{3}\right)+B\left(-\sqrt{3}\right)\left(\left(-\sqrt{3}\right)-\sqrt{3}\right)+C\left(-\sqrt{3}\right)\left(\left(-\sqrt{3}\right)+\sqrt{3}\right)\\ \\ 9+\sqrt{3} &=6B\\ \\ B &=\dfrac{9+\sqrt{3}}{6} \end{align} {/eq}

**Let** {eq}x-\sqrt{3}=0 \rightarrow x=\sqrt{3} {/eq}

{eq}\begin{align} \left(\sqrt{3}\right)^2-\left(\sqrt{3}\right)+6 &= A\left(\left(\sqrt{3}\right)+\sqrt{3}\right)\left(\left(\sqrt{3}\right)-\sqrt{3}\right)+B\left(\sqrt{3}\right)\left(\left(\sqrt{3}\right)-\sqrt{3}\right)+C\left(\sqrt{3}\right)\left(\left(\sqrt{3}\right)+\sqrt{3}\right)\\ \\ 9-\sqrt{3} &=6C\\ \\ C &=\dfrac{9-\sqrt{3}}{6} \end{align} {/eq}

Applying these values, we have:

{eq}\dfrac{\left(x^2-x+6\right)}{\left(x^3-3x\right)}=-\dfrac{2}{x}+\dfrac{9+\sqrt{3}}{6\left(x+\sqrt{3}\right)}+\dfrac{9-\sqrt{3}}{6\left(x-\sqrt{3}\right)} {/eq}

Integrating both sides:

{eq}\begin{align} \int \dfrac{(x^2-x+6)}{(x^3-3x)} dx &=\int \left [-\dfrac{2}{x}+\dfrac{9+\sqrt{3}}{6\left(x+\sqrt{3}\right)}+\dfrac{9-\sqrt{3}}{6\left(x-\sqrt{3}\right)} \right ]dx\\ \\ &=-\int \dfrac{2}{x}dx+\int \dfrac{9+\sqrt{3}}{6\left(x+\sqrt{3}\right)}dx+\int \dfrac{9-\sqrt{3}}{6\left(x-\sqrt{3}\right)}dx\\ \\ &=-2\int \dfrac{1}{x}dx+\dfrac{9+\sqrt{3}}{6}\int\dfrac{1}{\left(x+\sqrt{3}\right)}dx+ \dfrac{9-\sqrt{3}}{6}\int\dfrac{1}{\left(x-\sqrt{3}\right)}dx\\ \\ &=-2\ln \left|x\right|+\dfrac{9+\sqrt{3}}{6}\ln \left|x+\sqrt{3}\right|+\dfrac{9-\sqrt{3}}{6}\ln \left|x-\sqrt{3}\right|+C \end{align} {/eq}

{eq}\color{blue}{\int \dfrac{(x^2-x+6)}{(x^3-3x)} dx=-2\ln \left|x\right|+\dfrac{9+\sqrt{3}}{6}\ln \left|x+\sqrt{3}\right|+\dfrac{9-\sqrt{3}}{6}\ln \left|x-\sqrt{3}\right|+C} {/eq} ( where **C** is the constant of integration )

#### Learn more about this topic:

from AP Calculus AB & BC: Homework Help Resource

Chapter 13 / Lesson 13