# Question #804a1

Mar 30, 2017

$\int {x}^{3} / \left({x}^{2} - x - 2\right) \mathrm{dx} = {x}^{2} / 2 + x + \frac{1}{3} \ln \left\mid x + 1 \right\mid + \frac{8}{3} \ln \left\mid x - 2 \right\mid + C$

#### Explanation:

Consider the integrand function:

${x}^{3} / \left({x}^{2} - x - 2\right)$

Factorize the denominator:

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

Now add ans subtract $1$ to the numerator:

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

Use the factorization: $\left({x}^{3} + 1\right) = \left(x + 1\right) \left({x}^{2} - x + 1\right)$ so that we can simplify:

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

Now note that:

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

so :

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

and that using partial fractions:

$\frac{1}{\left(x + 1\right) \left(x - 2\right)} = \frac{A}{x + 1} + \frac{B}{x - 2}$

$\frac{1}{\left(x + 1\right) \left(x - 2\right)} = \frac{A x - 2 A + B x + B}{\left(x + 1\right) \left(x - 2\right)}$

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

$\left\{\begin{matrix}A = - \frac{1}{3} \\ B = \frac{1}{3}\end{matrix}\right.$

Substitute this in $\left(1\right)$ to have:

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

${x}^{3} / \left({x}^{2} - x - 2\right) = x + 1 + \frac{1}{3 \left(x + 1\right)} + \frac{8}{3 \left(x - 2\right)}$

We can now integrate:

$\int {x}^{3} / \left({x}^{2} - x - 2\right) \mathrm{dx} = {x}^{2} / 2 + x + \frac{1}{3} \ln \left\mid x + 1 \right\mid + \frac{8}{3} \ln \left\mid x - 2 \right\mid + C$
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