# How do you integrate  (x^2+8)/(x^2-5x+6) using partial fractions?

Sep 2, 2017

$\int \frac{{x}^{2} + 8}{{x}^{2} - 5 x + 6} \mathrm{dx} = x + 17 \ln \left\mid x - 3 \right\mid - 12 \ln \left\mid x - 2 \right\mid + C$

#### Explanation:

Factorize the denominator:

${x}^{2} - 5 x + 6 = \left(x - 2\right) \left(x - 3\right)$

Before performing partial fraction decomposition of the rational function, the numerator must have lower degree than the denominator, so split the function as:

$\frac{{x}^{2} + 8}{{x}^{2} - 5 x + 6} = \frac{{x}^{2} - 5 x + 6 + 5 x + 2}{{x}^{2} - 5 x + 6} = 1 + \frac{5 x + 2}{{x}^{2} - 5 x + 6}$

Now:

$\frac{5 x + 2}{{x}^{2} - 5 x + 6} = \frac{A}{x - 2} + \frac{B}{x - 3}$

$\frac{5 x + 2}{{x}^{2} - 5 x + 6} = \frac{A \left(x - 3\right) + B \left(x - 2\right)}{\left(x - 2\right) \left(x - 3\right)}$

$5 x + 2 = A x - 3 A + B x - 2 B$

$5 x + 2 = \left(A + B\right) x - \left(3 A + 2 B\right)$

$\left\{\begin{matrix}A + B = 5 \\ 3 A + 2 B = - 2\end{matrix}\right.$

$\left\{\begin{matrix}A = - 12 \\ B = 17\end{matrix}\right.$

So:

$\frac{5 x + 2}{{x}^{2} - 5 x + 6} = \frac{17}{x - 3} - \frac{12}{x - 2}$

$\frac{{x}^{2} + 8}{{x}^{2} - 5 x + 6} = 1 + \frac{17}{x - 3} - \frac{12}{x - 2}$

$\int \frac{{x}^{2} + 8}{{x}^{2} - 5 x + 6} \mathrm{dx} = \int \mathrm{dx} + 17 \int \frac{\mathrm{dx}}{x - 3} - 12 \int \frac{\mathrm{dx}}{x - 2}$

$\int \frac{{x}^{2} + 8}{{x}^{2} - 5 x + 6} \mathrm{dx} = x + 17 \ln \left\mid x - 3 \right\mid - 12 \ln \left\mid x - 2 \right\mid + C$

Sep 2, 2017

$x + 17 \ln | x - 3 | - 12 \ln | x - 2 | + C$

#### Explanation:

Dividing by the denominator leads to

 {x^2+8}/{x^2−5x+6} = 1+{5x+2}/{x^2−5x+6}

Since x^2−5x+6 = (x-3)(x-2) we can write

{5x+2}/{x^2−5x+6} = A/{x-3}+B/{x-2}

Multiplying both sides by x^2−5x+6 leads to

$5 x + 2 = A \left(x - 2\right) + B \left(x - 3\right)$

Substituting $x = 3$ and $x = 2$ in turn leads to the result $A = 17 , B = - 12$, so that

 {x^2+8}/{x^2−5x+6} = 1+17/{x-3}-12/{x-2}

Integrating this leads to the quoted result quickly!