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

Nov 15, 2015

$I = \frac{1}{50} \ln | 2 x + 1 | + \frac{87}{25} \ln | x - 2 | - \frac{29}{5 \left(x - 2\right)} + C$

#### Explanation:

$\left({x}^{2} - 4 x + 4\right) = {\left(x - 2\right)}^{2}$

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

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

$= \frac{A \left({x}^{2} - 4 x + 4\right) + B \left(2 {x}^{2} - 3 x - 2\right) + C \left(2 x + 1\right)}{\left(2 x + 1\right) {\left(x - 2\right)}^{2}} =$

$= \frac{A {x}^{2} - 4 A x + 4 A + 2 B {x}^{2} - 3 B x - 2 B + 2 C x + C}{\left(2 x + 1\right) {\left(x - 2\right)}^{2}} =$

$= \frac{{x}^{2} \left(A + 2 B\right) + x \left(- 4 A - 3 B + 2 C\right) + \left(4 A - 2 B + C\right)}{\left(2 x + 1\right) {\left(x - 2\right)}^{2}}$

$A + 2 B = 7 \implies A = 7 - 2 B$
$- 4 A - 3 B + 2 C = 1$
$4 A - 2 B + C = - 1$

$- 28 + 8 B - 3 B + 2 C = 1$
$28 - 8 B - 2 B + C = - 1$

$5 B + 2 C = 29$
$- 10 B + C = - 29$

$5 C = 29 \implies C = \frac{29}{5}$

$B = \frac{29 - 2 C}{5} = \frac{29}{5} - \frac{58}{25} = \frac{87}{25}$

$A = 7 - 2 B = 7 - \frac{174}{25} = \frac{1}{25}$

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

$I = \frac{1}{25} \int \frac{\mathrm{dx}}{2 x + 1} + \frac{87}{25} \int \frac{\mathrm{dx}}{x - 2} + \frac{29}{5} \int \frac{\mathrm{dx}}{x - 2} ^ 2$

$I = \frac{1}{50} \ln | 2 x + 1 | + \frac{87}{25} \ln | x - 2 | - \frac{29}{5 \left(x - 2\right)} + C$