# How do you integrate f(x)=(x^2+x)/((x^2-1)(x+3)(x-9)) using partial fractions?

Jun 21, 2018

$- \frac{1}{16} \cdot \ln | x + 3 | + \frac{3}{32} \ln | x - 9 | - \frac{1}{32} \ln | x - 1 | + C$

#### Explanation:

Converting your Integrand into partial fractions we get

$f \left(x\right) = - \frac{1}{12 \cdot \left(x - 1\right)} + \frac{3}{32 \left(x - 9\right)} - \frac{1}{16 \cdot \left(x + 3\right)}$
Integrating this we get the result above.

Aug 11, 2018

$\int \frac{{x}^{2} + x}{\left({x}^{2} - 1\right) \left(x + 3\right) \left(x - 9\right)} \mathrm{dx} = \frac{1}{22} \ln \left(| x - 1 |\right) - \frac{5}{44} \ln \left(| x + 3 |\right) + \frac{3}{44} \ln \left(| x - 9 |\right) + C$, $C \in \mathbb{R}$

#### Explanation:

$I = \int \frac{{x}^{2} + x}{\left({x}^{2} - 1\right) \left(x + 3\right) \left(x - 9\right)} \mathrm{dx}$

=int(xcancel((x+1)))/(cancel((x+1))(x-1)(x+3)(x-9)dx

Consider that $\frac{x}{\left(x - 1\right) \left(x + 3\right) \left(x - 9\right)} = \frac{A}{x - 1} + \frac{B}{x + 3} + \frac{C}{x - 9}$

$x = A \left(x + 3\right) \left(x - 9\right) + B \left(x - 1\right) \left(x - 9\right) + C \left(x - 1\right) \left(x + 3\right)$

$x = \left(A + B + C\right) {x}^{2} + \left(2 C - 6 A - 10 B\right) x - 27 A + 9 B - 3 C$

By identification :

$\left[\begin{matrix}A + B + C \\ - 6 A - 10 B + 2 C \\ - 27 A + 9 B - 3 C\end{matrix}\right] = \left[\begin{matrix}0 \\ 1 \\ 0\end{matrix}\right]$

Solving this system, you will find :

$A = \frac{1}{22}$, $B = - \frac{5}{44}$, $C = \frac{3}{44}$

So:

$I = \int \left(\frac{\frac{1}{22}}{x - 1} - \frac{\frac{5}{44}}{x + 3} + \frac{\frac{3}{44}}{x - 9}\right) \mathrm{dx}$

$= \frac{1}{22} \int \frac{1}{x - 1} \mathrm{dx} - \frac{5}{44} \int \frac{1}{x + 3} \mathrm{dx} + \frac{3}{44} \int \frac{1}{x - 9} \mathrm{dx}$

$= \frac{1}{22} \ln \left(| x - 1 |\right) - \frac{5}{44} \ln \left(| x + 3 |\right) + \frac{3}{44} \ln \left(| x - 9 |\right) + C$, $C \in \mathbb{R}$

\0/ Here's our answer !