# How do you integrate int (2x+1)/((x+3)(x-2)(x-7))  using partial fractions?

Apr 21, 2016

Perform a lot of boring algebra to end up with $- \frac{1}{10} \ln \left\mid x + 3 \right\mid - \frac{1}{5} \ln \left\mid x - 2 \right\mid + \frac{3}{10} \ln \left\mid x - 7 \right\mid + C$.

#### Explanation:

Lucky for us, the denominator in the integral is nice and factored for us, so our partial fraction decomposition will be of the form $\frac{A}{x + 3} + \frac{B}{x - 2} + \frac{C}{x - 7}$.

Start by breaking $\frac{2 x + 1}{\left(x + 3\right) \left(x - 2\right) \left(x - 7\right)}$ into partial fractions:
$\frac{2 x + 1}{\left(x + 3\right) \left(x - 2\right) \left(x - 7\right)} = \frac{A}{x + 3} + \frac{B}{x - 2} + \frac{C}{x - 7}$

Multiplying through by $\left(x + 3\right) \left(x - 2\right) \left(x - 7\right)$ gives:
$2 x + 1 = A \left(x - 2\right) \left(x - 7\right) + B \left(x + 3\right) \left(x - 7\right) + C \left(x + 3\right) \left(x - 2\right)$

That equation is looking pretty nasty, but note that if we set $x = 2$, we get:
$2 \left(2\right) + 1 = A \left(\left(2\right) - 2\right) \left(\left(2\right) - 7\right) + B \left(\left(2\right) + 3\right) \left(\left(2\right) - 7\right) + C \left(\left(2\right) + 3\right) \left(\left(2\right) - 2\right)$
$5 = A \left(0\right) \left(- 5\right) + B \left(5\right) \left(- 5\right) + C \left(5\right) \left(0\right)$
$5 = - 25 B$
$B = - \frac{1}{5}$

Likewise, set $x = 7$ to get:
$2 \left(7\right) + 1 = A \left(\left(7\right) - 2\right) \left(\left(7\right) - 7\right) + B \left(\left(7\right) + 3\right) \left(\left(7\right) - 7\right) + C \left(\left(7\right) + 3\right) \left(\left(7\right) - 2\right)$
$15 = A \left(5\right) \left(0\right) + B \left(10\right) \left(0\right) + C \left(10\right) \left(5\right)$
$15 = 50 C$
$C = \frac{15}{50} = \frac{3}{10}$

And finally, set $x = - 3$ to find $A$:
$2 \left(- 3\right) + 1 = A \left(\left(- 3\right) - 2\right) \left(\left(- 3\right) - 7\right) + B \left(\left(- 3\right) + 3\right) \left(\left(- 3\right) - 7\right) + C \left(\left(- 3\right) + 3\right) \left(\left(- 3\right) - 2\right)$
$- 5 = A \left(- 5\right) \left(- 10\right) + B \left(0\right) \left(- 10\right) + C \left(0\right) \left(- 5\right)$
$- 5 = 50 A$
$A = - \frac{5}{50} = - \frac{1}{10}$

So we have $A = - \frac{1}{10}$, $B = - \frac{1}{5}$, and $C = \frac{3}{10}$. Plugging these back into our original equation, we see:
$\frac{2 x + 1}{\left(x + 3\right) \left(x - 2\right) \left(x - 7\right)} = \frac{- \frac{1}{10}}{x + 3} + \frac{- \frac{1}{5}}{x - 2} + \frac{\frac{3}{10}}{x - 7}$
$\textcolor{w h i t e}{X X} = - \frac{1}{10 \left(x + 3\right)} - \frac{1}{5 \left(x - 2\right)} + \frac{3}{10 \left(x - 7\right)}$

Now all that's left is to integrate this:
$\int \frac{2 x + 1}{\left(x + 3\right) \left(x - 2\right) \left(x - 7\right)} \mathrm{dx} = \int - \frac{1}{10 \left(x + 3\right)} - \frac{1}{5 \left(x - 2\right)} + \frac{3}{10 \left(x - 7\right)} \mathrm{dx}$
$\textcolor{w h i t e}{X X} = \int - \frac{1}{10 \left(x + 3\right)} \mathrm{dx} - \int \frac{1}{5 \left(x - 2\right)} \mathrm{dx} + \int \frac{3}{10 \left(x - 7\right)} \mathrm{dx}$
$\textcolor{w h i t e}{X X} = - \frac{1}{10} \int \frac{1}{x + 3} \mathrm{dx} - \frac{1}{5} \int \frac{1}{x - 2} \mathrm{dx} + \frac{3}{10} \int \frac{1}{x - 7} \mathrm{dx}$
$\textcolor{w h i t e}{X X} = - \frac{1}{10} \ln \left\mid x + 3 \right\mid - \frac{1}{5} \ln \left\mid x - 2 \right\mid + \frac{3}{10} \ln \left\mid x - 7 \right\mid + C$

And we're done.