# How do you find int (3x^2 - 10) / (x^2-4x+4) dx using partial fractions?

Feb 27, 2016

$3 x + 12 \ln \left\mid x - 2 \right\mid - \frac{2}{x - 2} + C$

#### Explanation:

First, since the degrees of the numerator and denominator are equal, use polynomial long division to rewrite the expression:

$\frac{3 {x}^{2} - 10}{{x}^{2} - 4 x + 4} = 3 + \frac{12 x - 22}{{x}^{2} - 4 x + 4}$

Now, perform partial fraction decomposition on $\frac{12 x - 22}{{x}^{2} - 4 x + 4}$, recognizing that ${x}^{2} - 4 x + 4 = {\left(x - 2\right)}^{2}$.

$\frac{12 x - 22}{{\left(x - 2\right)}^{2}} = \frac{A}{x - 2} + \frac{B}{x - 2} ^ 2$

Note that since the term is squared, it will be repeated.

Multiply both sides by ${\left(x - 2\right)}^{2}$ to see that

$12 x - 22 = A \left(x - 2\right) + B$

When we set $x = 2$, we see that

$12 \left(2\right) - 22 = A \left(0\right) + B$

$2 = B$

Arbitrarily, set $x = 3$ to solve for $A$, recalling that $B = 2$:

$12 \left(3\right) - 22 = A \left(1\right) + 2$

$A = 12$

Thus,

$\frac{3 {x}^{2} - 10}{{x}^{2} - 4 x + 4} = 3 + \frac{12}{x - 2} + \frac{2}{x - 2} ^ 2$

Now, we can integrate more simply:

$\int 3 + \frac{12}{x - 2} + \frac{2}{x - 2} ^ 2 \mathrm{dx}$

$= 3 x + 12 \ln \left\mid x - 2 \right\mid - \frac{2}{x - 2} + C$