# How do you integrate int (4x^3 - 4x^2 - 16x + 7) / ((x + 1) (x -2)) using partial fractions?

Oct 15, 2017

$\int \setminus \frac{4 {x}^{3} - 4 {x}^{2} - 16 x + 7}{\left(x + 1\right) \left(x - 2\right)} \setminus \mathrm{dx} = 2 {x}^{2} - 5 \ln | x + 1 | - 3 \ln | x - 2 | + C$

#### Explanation:

We seek:

$I = \int \setminus \frac{4 {x}^{3} - 4 {x}^{2} - 16 x + 7}{\left(x + 1\right) \left(x - 2\right)} \setminus \mathrm{dx}$

First, we note that the order of the polynomial of the numerator is higher than that of the denominator. We therefore have the algebraic equivalent of a "top-heavy" fraction, and so we must first use algebraic long division:

 {: ( , , ul(+4x), ul(" "), ul(" "), ul(" "), ), ( x^2-x-2, ")", +4x^3, -4x^2, -16x, + 7, ), ( , , ul(+4x^3), ul(-4x^2), ul(-8x), ul(" "), -), ( , , , , -8x, +7, ) :}

So, we find:

$\frac{4 {x}^{3} - 4 {x}^{2} - 16 x + 7}{{x}^{2} - x - 2} \equiv 4 x + \frac{- 8 x + 7}{{x}^{2} - x - 2}$

We can now decompose the second term in to partial fractions:

$\frac{- 8 x + 7}{{x}^{2} - x - 2} \equiv \frac{- 8 x + 7}{\left(x + 1\right) \left(x - 2\right)}$
$\text{ } = \frac{A}{x + 1} + \frac{B}{x - 2}$
$\text{ } = \frac{A \left(x - 2\right) + B \left(x + 1\right)}{\left(x + 1\right) \left(x - 2\right)}$

$- 8 x + 7 = A \left(x - 2\right) + B \left(x + 1\right)$

Where $A , B$ are constants that are to be determined. We can find them by substitutions (In practice we do this via the "cover up" method:

Put $x = - 1 \implies 8 + 7 = A \left(- 3\right) \implies A = - 5$
Put $x = + 2 \implies - 16 + 7 = 3 B \implies B = - 3$

So we can now write:

$I = \int \setminus 4 x - \frac{5}{x + 1} - \frac{3}{x - 2} \setminus \mathrm{dx}$

Which now consists of standard integral so we integrate to get:

$I = 2 {x}^{2} - 5 \ln | x + 1 | - 3 \ln | x - 2 | + C$