# How do you integrate (x^3 - 5x + 2) / (x^2 - 8x + 15) using partial fractions?

Apr 8, 2018

The answer is $= {x}^{2} / 2 + 8 x + 51 \ln \left(| x - 5 |\right) - 7 \ln \left(| x - 3 |\right) + C$

#### Explanation:

Perform a polynomial division first

$\frac{{x}^{3} - 5 x + 2}{{x}^{2} - 8 x + 15} = \left(x + 8\right) + \frac{44 x - 118}{\left(x - 5\right) \left(x - 3\right)}$

Perform the decomposition into partial fractions

$\frac{44 x - 118}{\left(x - 5\right) \left(x - 3\right)} = \frac{A}{x - 5} + \frac{B}{x - 3}$

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

The denominators are the same, compare the numerators

$44 x - 118 = \left(A \left(x - 3\right) + B \left(x - 5\right)\right)$

Let $x = 5$, $\implies$, $102 = 2 A$, $\implies$, $A = 51$

Let $x = 3$, $\implies$, $14 = - 2 B$, $\implies$, $B = - 7$

Therefore,

$\frac{44 x - 118}{\left(x - 5\right) \left(x - 3\right)} = \frac{51}{x - 5} - \frac{7}{x - 3}$

and

$\frac{{x}^{3} - 5 x + 2}{{x}^{2} - 8 x + 15} = \left(x + 8\right) + \frac{51}{x - 5} - \frac{7}{x - 3}$

$\int \frac{\left({x}^{3} - 5 x + 2\right) \mathrm{dx}}{{x}^{2} - 8 x + 15} = \int \left(x + 8\right) \mathrm{dx} + \int \frac{51 \mathrm{dx}}{x - 5} - \int \frac{7 \mathrm{dx}}{x - 3}$

$= {x}^{2} / 2 + 8 x + 51 \ln \left(| x - 5 |\right) - 7 \ln \left(| x - 3 |\right) + C$

Apr 8, 2018

$\frac{1}{2} {x}^{2} + 8 x + 22 \ln | \left({x}^{2} - 8 x + 15\right) | + 29 \ln | \frac{x - 5}{x - 3} | + C .$

#### Explanation:

Observe that the integrand is an improper rational function,

i.e., the degree of the Nr. poly. is more than that of the Dr.

So, first of all we will have to make it proper. Usually, the

long division is performed, but, let us proceed as under :

${x}^{3} - 5 x + 2 = \textcolor{red}{x} \left({x}^{2} \textcolor{red}{- 8 x + 15}\right) + \textcolor{red}{8 {x}^{2} - 15 x} - 5 x + 2$,

$= x \left({x}^{2} - 8 x + 15\right) + 8 {x}^{2} - 20 x + 2$,

$= x \left({x}^{2} - 8 x + 15\right) + 8 \left({x}^{2} - 8 x + 15\right) + 64 x - 120 - 20 x + 2$,

$\therefore {x}^{3} - 5 x + 2 = \left(x + 8\right) \left({x}^{2} - 8 x + 15\right) + 44 x - 118$.

$\therefore \frac{{x}^{3} - 5 x + 2}{{x}^{2} - 8 x + 15} = \left(x + 8\right) + \frac{44 x - 118}{{x}^{2} - 8 x + 15}$.

Since, $\frac{d}{\mathrm{dx}} \left({x}^{2} - 8 x + 15\right) = 2 x - 8$, we write,

$\left(44 x - 118\right) = 22 \left(2 x - 8\right) + 176 - 118 = 22 \left(2 x - 8\right) + 58$.

$\therefore \int \frac{{x}^{3} - 5 x + 2}{{x}^{2} - 8 x + 15} \mathrm{dx}$,

$= \int \left[\left(x + 8\right) + \frac{22 \frac{d}{\mathrm{dx}} \left({x}^{2} - 8 x + 15\right) + 58}{{x}^{2} - 8 x + 15}\right] \mathrm{dx}$,

$= \frac{1}{2} {x}^{2} + 8 x + 22 \int \frac{\frac{d}{\mathrm{dx}} \left({x}^{2} - 8 x + 15\right)}{{x}^{2} - 8 x + 15} \mathrm{dx}$

$+ 58 \int \frac{1}{{\left(x - 4\right)}^{2} - 1} \mathrm{dx}$,

$= \frac{1}{2} {x}^{2} + 8 x + 22 \ln | \left({x}^{2} - 8 x + 15\right) | + 58 \cdot \frac{1}{2} \ln | \frac{\left(x - 4\right) - 1}{\left(x - 4\right) + 1} |$

$= \frac{1}{2} {x}^{2} + 8 x + 22 \ln | \left({x}^{2} - 8 x + 15\right) | + 29 \ln | \frac{x - 5}{x - 3} | + C .$

Enjoy Maths.!

N.B. : It will be an interesting exercise for the

Questioner to verify that this Answer is the

same as the one submitted by Respected Narad T.!