# How do you find the integral of (x^3-4x-10)/(x^2-x-6) from 0 to 1?

Mar 19, 2018

${\int}_{0}^{1} \frac{{x}^{3} - 4 x - 10}{{x}^{2} - x - 6} \mathrm{dx} = \frac{3}{2} + \ln \left(\frac{3}{2}\right)$

#### Explanation:

Solve first the indefinite integral, noting that:

${x}^{2} - x - 6 = \left(x - 3\right) \left(x + 2\right)$

${x}^{3} - 4 x - 10 = x \left({x}^{2} - 4\right) - 10 = x \left(x - 2\right) \left(x + 2\right) - 10$

So:

$\int \frac{{x}^{3} - 4 x - 10}{{x}^{2} - x - 6} \mathrm{dx} = \int \frac{x \left(x - 2\right) \left(x + 2\right) - 10}{\left(x - 3\right) \left(x + 2\right)} \mathrm{dx}$

Simplify and use the linearity of the integral:

$\int \frac{{x}^{3} - 4 x - 10}{{x}^{2} - x - 6} \mathrm{dx} = \int \frac{x \left(x - 2\right)}{x - 3} \mathrm{dx} - 10 \int \frac{\mathrm{dx}}{\left(x - 3\right) \left(x + 2\right)}$

Solve the integrals separately. In the first substitute $\left(x - 3\right) = u$:

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

$\int \frac{x \left(x - 2\right)}{x - 3} \mathrm{dx} = \int \frac{{u}^{2} + 4 u + 3}{u} \mathrm{du}$

and always using linearity:

$\int \frac{x \left(x - 2\right)}{x - 3} \mathrm{dx} = \int u \mathrm{du} + 4 \int \mathrm{du} + 3 \int \frac{\mathrm{du}}{u}$

$\int \frac{x \left(x - 2\right)}{x - 3} \mathrm{dx} = {u}^{2} / 2 + 4 u + 3 \ln \left\mid u \right\mid$

and undoing the substitution:

$\int \frac{x \left(x - 2\right)}{x + 3} \mathrm{dx} = {\left(x - 3\right)}^{2} / 2 + 4 \left(x - 3\right) + 3 \ln \left\mid x - 3 \right\mid + C$

Solve the second integral with partial fraction decomposition:

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

$1 = A \left(x + 2\right) + B \left(x - 3\right)$

$1 = \left(A + B\right) x + \left(2 A - 3 B\right)$

$\left\{\begin{matrix}A + B = 0 \\ 2 A - 3 B = 1\end{matrix}\right.$

$\left\{\begin{matrix}A = \frac{1}{5} \\ B = - \frac{1}{5}\end{matrix}\right.$

So:

$\int \frac{\mathrm{dx}}{\left(x - 3\right) \left(x + 2\right)} = \frac{1}{5} \int \frac{\mathrm{dx}}{x - 3} - \frac{1}{5} \int \frac{\mathrm{dx}}{x + 2}$

$\int \frac{\mathrm{dx}}{\left(x - 3\right) \left(x + 2\right)} = \frac{1}{5} \ln \left\mid x - 3 \right\mid - \frac{1}{5} \ln \left\mid x + 2 \right\mid + C$

Put together the partial results:

$\int \frac{{x}^{3} - 4 x - 10}{{x}^{2} - x - 6} \mathrm{dx} = {\left(x - 3\right)}^{2} / 2 + 4 \left(x - 3\right) + 3 \ln \left\mid x - 3 \right\mid - 2 \ln \left\mid x - 3 \right\mid + 2 \ln \left\mid x + 2 \right\mid + C$

Simplifying and absorbing the constants in $C$:

$\int \frac{{x}^{3} - 4 x - 10}{{x}^{2} - x - 6} \mathrm{dx} = {x}^{2} / 2 + x + \ln \left\mid x - 3 \right\mid + 2 \ln \left\mid x + 2 \right\mid + C$

Now:

${\int}_{0}^{1} \frac{{x}^{3} - 4 x - 10}{{x}^{2} - x - 6} \mathrm{dx} = \frac{1}{2} + 1 + \ln 2 + 2 \ln 3 - \ln 3 - 2 \ln 2$

${\int}_{0}^{1} \frac{{x}^{3} - 4 x - 10}{{x}^{2} - x - 6} \mathrm{dx} = \frac{3}{2} - \ln 2 + \ln 3$

${\int}_{0}^{1} \frac{{x}^{3} - 4 x - 10}{{x}^{2} - x - 6} \mathrm{dx} = \frac{3}{2} + \ln \left(\frac{3}{2}\right)$