# How do you integrate int (3x^3+2x^2-7x-6)/(x^2-4) dx using partial fractions?

Nov 10, 2016

The integral is $y = \frac{3}{2} {x}^{2} + 2 x + 2 \ln | x + 2 | + 3 \ln | x - 2 | + C$

#### Explanation:

Start by using long division to divide. We want the degree of the numerator to be less than the degree of the denominator.

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So, the quotient is $3 x + 2$ with a remainder of $5 x + 2$, which can be written as $3 x + 2 + \frac{5 x + 2}{{x}^{2} - 4}$.

We can now write $\frac{5 x + 2}{{x}^{2} - 4}$ in partial fractions.

$\frac{A}{x + 2} + \frac{B}{x - 2} = \frac{5 x + 2}{\left(x + 2\right) \left(x - 2\right)}$

(A(x- 2)) + B(x + 2)) = 5x + 2

$A x - 2 A + B x + 2 B = 5 x + 2$

$\left(A + B\right) x + \left(2 B - 2 A\right) = 5 x + 2$

So, $A + B = 5$ and $2 B - 2 A = 2$.

$B = 5 - A$

$2 \left(5 - A\right) - 2 A = 2$

$10 - 2 A - 2 A = 2$

$- 4 A = - 8$

$A = 2$

$\therefore 2 + B = 5$

$B = 3$

So, the partial fraction decomposition is $\frac{2}{x + 2} + \frac{3}{x - 2}$.

We rewrite the integral like this:

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

Integrate using the rule $\int \left(\frac{1}{x}\right) \mathrm{dx} = \ln x + C$.

$= \frac{3}{2} {x}^{2} + 2 x + 2 \ln | x + 2 | + 3 \ln | x - 2 | + C$

You can check the answer by differentiating. You will get the initial problem.

Hopefully this helps!