Jun 21, 2018

$\frac{5}{3} \ln | \left(x - 3\right) | - \frac{3}{2} \ln | \left(x - 2\right) | - \frac{1}{6} \ln | x | + C , \mathmr{and} ,$

$\ln | {\left(x - 3\right)}^{\frac{5}{3}} / \left\{{\left(x - 2\right)}^{\frac{3}{2}} {x}^{\frac{1}{6}}\right\} | + C$.

#### Explanation:

Suppose that, $I = \int \frac{2 x - 1}{x \left(x - 2\right) \left(x - 3\right)} \mathrm{dx}$.

$\therefore I = \int \left\{\frac{2 x}{x \left(x - 2\right) \left(x - 3\right)} + \frac{- 1}{x \left(x - 2\right) \left(x - 3\right)}\right\} \mathrm{dx}$,

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

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

$= 2 \int \frac{1}{\left(x - 2\right) \left(x - 3\right)} \mathrm{dx} + \int \frac{1}{x \left(x - 2\right)} \mathrm{dx} - \frac{1}{x \left(x - 3\right)} \mathrm{dx}$,

$= 2 \int \frac{\left(x - 2\right) - \left(x - 3\right)}{\left(x - 2\right) \left(x - 3\right)} \mathrm{dx} + \frac{1}{2} \int \frac{x - \left(x - 2\right)}{x \left(x - 2\right)} \mathrm{dx}$

$- \frac{1}{3} \int \frac{x - \left(x - 3\right)}{x \left(x - 3\right)} \mathrm{dx}$,

$= 2 \int \left\{\frac{1}{x - 3} - \frac{1}{x - 2}\right\} \mathrm{dx} + \frac{1}{2} \int \left\{\frac{1}{x - 2} - \frac{1}{x}\right\} \mathrm{dx}$

$- \frac{1}{3} \int \left\{\frac{1}{x - 3} - \frac{1}{x}\right\} \mathrm{dx}$,

$= 2 \ln | \left(x - 3\right) | - 2 \ln | \left(x - 2\right) | + \frac{1}{2} \ln | \left(x - 2\right) | - \frac{1}{2} \ln | x | - \frac{1}{3} \ln | \left(x - 3\right) | + \frac{1}{3} \ln | x |$,

$= \left(2 - \frac{1}{3}\right) \ln | \left(x - 3\right) | - \left(2 - \frac{1}{2}\right) \ln | \left(x - 2\right) | - \left(\frac{1}{2} - \frac{1}{3}\right) \ln | x |$.

$\Rightarrow I = \frac{5}{3} \ln | \left(x - 3\right) | - \frac{3}{2} \ln | \left(x - 2\right) | - \frac{1}{6} \ln | x | + C , \mathmr{and} ,$

$I = \ln | {\left(x - 3\right)}^{\frac{5}{3}} / \left\{{\left(x - 2\right)}^{\frac{3}{2}} {x}^{\frac{1}{6}}\right\} | + C$.

Jun 21, 2018

$\frac{5}{3} \ln | x - 3 | - \frac{3}{2} \cdot \ln | x - 2 | - \frac{1}{6} \cdot \ln | x | + C$

#### Explanation:

Converting your Integrand in partial fractions we get

$\frac{2 x - 1}{x \cdot \left(x - 2\right) \left(x - 3\right)} = \frac{5}{3 \cdot \left(x - 3\right)} - \frac{3}{2 \cdot \left(x - 2\right)} - \frac{1}{6 \cdot x}$

The result is to be seen above.