# How do you integrate int (x+4) /((x+1)(x - 2)²) using partial fractions?

Jun 30, 2016

Solution is

$\frac{1}{3} \cdot \ln | x + 1 | - \frac{1}{3} \cdot \ln | x - 2 | - \frac{2}{x - 2}$

#### Explanation:

This could become quite lengthy, just warning you.

$\frac{x + 4}{\left(x + 1\right) {\left(x - 2\right)}^{2}} = \frac{A}{x + 1} + \frac{B}{x - 2} + \frac{C}{{\left(x - 2\right)}^{2}}$

where A,B and C are arbitrary constants.

Multiplying appropriately to get common denominators we get:

$x + 4 = A {\left(x - 2\right)}^{2} + B \left(x + 1\right) \left(x - 2\right) + C \left(x + 1\right)$

There are a number of methods which can be used to determine the values of the constants, we shall set the value of x such that certain brackets evaluate to zero:

$x = 2$

$\therefore 2 + 4 = 3 \cdot C \implies C = 2$

$x = - 1$

$\therefore - 1 + 4 = 9 \cdot A \implies A = \frac{1}{3}$

Now we know A and C, try at x = 0:

$0 + 4 = \frac{1}{3} \cdot {\left(- 2\right)}^{2} - 2 \cdot B + 2$

$4 = \frac{4}{3} + 2 - 2 \cdot B \implies B = - \frac{1}{3}$

So the integral has become:

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

For the first part, let $u = x + 1 \implies \mathrm{du} = \mathrm{dx}$
for the second and third, let $v = x - 2 \implies \mathrm{dv} = \mathrm{dx}$

$\frac{1}{3} \cdot \int \frac{\mathrm{du}}{u} - \frac{1}{3} \cdot \int \frac{\mathrm{dv}}{v} + 2 \cdot \int \frac{\mathrm{dv}}{{v}^{2}}$

$= \frac{1}{3} \cdot \ln | x + 1 | - \frac{1}{3} \cdot \ln | x - 2 | - \frac{2}{x - 2}$