$=\int \left\{ \left( x+3 \right)+\frac{7x-6}{\left( x-1 \right)\left( x-2 \right)} \right\}dx$

$=\frac{{{x}^{2}}}{2}+3x+\int \frac{7x-6}{(x-1)(x-2)}dx$

$=\frac{{{x}^{2}}}{2}+3x+{{I}_{1}}……(1)$

Where,

${{I}_{1}}=\int \frac{7x-6}{\left( x-1 \right)\left( x-2 \right)}dx$

Putting $\frac{7x-6}{\left( x-1 \right)\left( x-2 \right)}=\frac{A}{x-1}+\frac{B}{x-2}…..(2)$

$A\left( x-2 \right)+B(x-1)=7x-6$

Now put, $x-2=0$

Therefore, $x=2$

$A\left( 0 \right)+B(2-1)=7\times 2-6$

$B=8$

Now put, $x-1=0$

Therefore, $x=1$

$A(1-2)+B(0)=7-6=1$

$A=-1$

Now from equation (2) we get,

$\frac{7x-6}{(x+1)(x-2)}=\frac{-1}{x-1}+\frac{8}{x-2}$

$I_{1}^{{}}=\int \frac{7x-6}{\left( x-1 \right)\left( x-2 \right)}dx=-\int \frac{1}{x-1}dx+8\int \frac{1}{x-2}dx$

$=-\log \left| x-1 \right|+8\log \left| x-2 \right|+c$

Now from equation (1) we get,

$I=\frac{{{x}^{2}}}{2}+3x-\log \left| x-1 \right|+8\log \left| x-2 \right|+c$