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

Sep 22, 2017

Given: $\int \frac{6 {x}^{2} + 7 x - 6}{\left({x}^{2} - 4\right) \left(x + 2\right)} \mathrm{dx}$

Decompose the integrand into partial fractions:

$\frac{6 {x}^{2} + 7 x - 6}{\left(x - 2\right) {\left(x + 2\right)}^{2}} = A \frac{1}{x - 2} + B \frac{1}{x + 2} ^ 2 + C \frac{1}{x + 2}$

Multiply both sides by $\left(x - 2\right) {\left(x + 2\right)}^{2}$:

$6 {x}^{2} + 7 x - 6 = A {\left(x + 2\right)}^{2} + B \left(x - 2\right) + C \left(x - 2\right) \left(x + 2\right)$

Find the value of A by letting $x = 2$:

$6 {\left(2\right)}^{2} + 7 \left(2\right) - 6 = A {\left(\left(2\right) + 2\right)}^{2} + B \left(2 - 2\right) + C \left(2 - 2\right) \left(2 + 2\right)$

$32 = A {\left(4\right)}^{2}$

$A = 2$

Find the value of B by letting $x = - 2$

$6 {\left(- 2\right)}^{2} + 7 \left(- 2\right) - 6 = 2 {\left(- 2 + 2\right)}^{2} + B \left(- 2 - 2\right) + C \left(- 2 - 2\right) \left(- 2 + 2\right)$

$4 = B \left(- 2 - 2\right)$

$B = - 1$

Find the value of C by letting $x = 0$

$6 {\left(0\right)}^{2} + 7 \left(0\right) - 6 = 2 {\left(0 + 2\right)}^{2} - 1 \left(0 - 2\right) + C \left(0 - 2\right) \left(0 + 2\right)$

$- 6 = 8 + 2 - 4 C$

$C = 4$

Here is the decomposition:

$\frac{6 {x}^{2} + 7 x - 6}{\left(x - 2\right) {\left(x + 2\right)}^{2}} = 2 \frac{1}{x - 2} - \frac{1}{x + 2} ^ 2 + 4 \frac{1}{x + 2}$

Therefore, we can write the following equation:

$\int \frac{6 {x}^{2} + 7 x - 6}{\left({x}^{2} - 4\right) \left(x + 2\right)} \mathrm{dx} = 2 \int \frac{1}{x - 2} \mathrm{dx} - \int \frac{1}{x + 2} ^ 2 \mathrm{dx} + 4 \int \frac{1}{x + 2} \mathrm{dx}$

The integrals on the right are trival:

$\int \frac{6 {x}^{2} + 7 x - 6}{\left({x}^{2} - 4\right) \left(x + 2\right)} \mathrm{dx} = 2 \ln | x - 2 | + \frac{1}{x + 2} + 4 \ln | x + 2 | + C$

Sep 22, 2017

2 ln|x-2| +4 ln |x+2| + 1/(x+2) +C

#### Explanation:

First of all the denominator can be put in a different shape as $\left(x - 2\right) {\left(x + 2\right)}^{2}$. The partial fractions can thus be put as $\frac{A}{x - 2} + \frac{B x}{x + 2} ^ 2 + \frac{C}{x + 2}$

The numerator would thus become $A {\left(x + 2\right)}^{2} + B x \left(x - 2\right) + C \left({x}^{2} - 4\right)$

On simplification this adds up to $A \left({x}^{2} + 4 x + 4\right) + B {x}^{2} - 2 B x + C {x}^{2} - 4 C$ or, # (A+B+C)x^2 + (4A-2B)x +4A-4C.

Now comparing like terms with $6 {x}^{2} + 7 x - 6$, it would be A+B+C=6, 4A-2B= 7 and4A-4C= -6. Solving these three equations, it would be A=2, B=$\frac{1}{2}$ and C=$\frac{7}{2}$

The integral thus becomes $\int \frac{2}{x - 2} \mathrm{dx} + \frac{1}{2} \int \frac{x}{x + 2} ^ 2 \mathrm{dx} + \frac{7}{2} \int \frac{1}{x + 2} \mathrm{dx}$

Or, $\int \frac{2}{x - 2} \mathrm{dx} + \frac{1}{2} \int \frac{x + 2 - 2}{x + 2} ^ 2 \mathrm{dx} + \frac{7}{2} \int \frac{1}{x + 2} \mathrm{dx}$

0r, $\int \frac{2}{x - 2} \mathrm{dx} + \int \frac{4}{x + 2} \mathrm{dx} - \int \frac{1}{x + 2} ^ 2 \mathrm{dx}$

2 ln|x-2| +4 ln |x+2|+ 1/(x+2) +C Ans