$\int \frac{3\mathrm{x}-1}{(\mathrm{x}-1)(\mathrm{x}-2)(\mathrm{x}-3)}\mathrm{dx}$ is equal to

Let $\frac{3\mathrm{x}-1}{(\mathrm{x}-1)(\mathrm{x}-2)(\mathrm{x}-3)}=\frac{\mathrm{A}}{(\mathrm{x}-1)}+\frac{\mathrm{B}}{(\mathrm{x}-2)}+\frac{\mathrm{C}}{(\mathrm{x}-3)}$
$\Rightarrow \frac{3\mathrm{x}-1}{(\mathrm{x}-1)(\mathrm{x}-2)(\mathrm{x}-3)}=\frac{\mathrm{A}(\mathrm{x}-2)(\mathrm{x}-3)+\mathrm{B}(\mathrm{x}-1)(\mathrm{x}-3)+\mathrm{C}(\mathrm{x}-1)(\mathrm{x}-2)}{(\mathrm{x}-1)(\mathrm{x}-2)(\mathrm{x}-3)}$
$\begin{array}{l}\Rightarrow 3\mathrm{x}-1=\mathrm{A}\left[{\mathrm{x}}^2-5\mathrm{x}+6\right]+\mathrm{B}\left[{\mathrm{x}}^2-4\mathrm{x}+3\right]+\mathrm{C}\left[{\mathrm{x}}^2-3\mathrm{x}+2\right]\\ \Rightarrow 3\mathrm{x}-1={\mathrm{x}}^2(\mathrm{A}+\mathrm{B}+\mathrm{C})+\mathrm{x}(-5\mathrm{A}-4\mathrm{B}-3\mathrm{C})+(6\mathrm{A}+3\mathrm{B}+2\mathrm{C})\end{array}$
On equating the coefficients of x² ,x and constant term on both sides, we get
$$\begin{gathered} \begin{array}{l} \mathrm{A}+\mathrm{B}+\mathrm{C}=0 .....\left(\mathrm{i}\right)\\ -5\mathrm{A}-4\mathrm{B}-3\mathrm{C}=3 .....\left(\mathrm{ii}\right) \\ \mathrm{and} 6\mathrm{A}+3\mathrm{B}+2\mathrm{C}=-1 .....\left(\mathrm{iii}\right)\end{array} \end{gathered}$$
From Eq. (i), we get A = -(B + C)
On putting the value of A in Eqs. (ii) and (iii), we get
     $-5\{-(\mathrm{B}+\mathrm{C}\left)\right\}-4\mathrm{B}-3\mathrm{C}=3$
$\begin{array}{l}\Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}5\mathrm{B}+5\mathrm{C}-4\mathrm{B}-3\mathrm{C}=3\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}6+2\mathrm{C}=3 .....\left(\mathrm{iv}\right)\\ \text{ and }\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}6\{-(\mathrm{B}+\mathrm{C}\left)\right\}+3\mathrm{B}+2\mathrm{C}=-1\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}-6\mathrm{B}-6\mathrm{C}+3\mathrm{B}+2\mathrm{C}=-1\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}-3\mathrm{B}-4\mathrm{C}=-1 ......\left(\mathrm{v}\right)\end{array}$
On solving Eqs. (iv) and (v), we get C = 4 On putting the value of C in Eq. (iv), we get
                         $\mathrm{B}+2\times 4=3$
$\Rightarrow \mathrm{B}=-5$
On putting the value of Band C in Eq. (i), we get
$$\begin{gathered} \mathrm{A}+(-5)+4=0\Rightarrow \mathrm{A}=1 \\ \therefore \mathrm{A}=1,\mathrm{B}=-5,\mathrm{C}=4 \end{gathered}$$
Now, 
$$\begin{gathered} \int \frac{3\mathrm{x}-1}{(\mathrm{x}-1)(\mathrm{x}-2)(\mathrm{x}-3)}\mathrm{dx}=\int \left[\frac{\mathrm{A}}{(\mathrm{x}-1)}+\frac{\mathrm{B}}{(\mathrm{x}-2)}+\frac{\mathrm{C}}{(\mathrm{x}-3)}\right]\mathrm{dx} \\ \begin{array}{r}=\int \frac{1}{(\mathrm{x}-1)}\mathrm{dx}+\int \frac{(-5)}{(\mathrm{x}-2)}\mathrm{dx}+\int \frac{4}{(\mathrm{x}-3)}\mathrm{dx}\\ =\log |\mathrm{x}-1|-5\log |\mathrm{x}-2|+4|\log |\mathrm{x}-3\mid +\mathrm{C}\end{array} \\ \left[\because \frac{1}{\mathrm{x}}\mathrm{dx}=\log \mathrm{x}\right] \end{gathered}$$