$\int \frac{x^{2}+1}{x^{2}-5 x+6} d x$
Consider $I=\int \frac{x^{2}+1}{x^{2}-5 x+6} d x$
Expressing the integral $\int \frac{\mathrm{P}(\mathrm{x})}{\mathrm{ax}^{2}+\mathrm{bx}+\mathrm{c}} \mathrm{dx}=\int \mathrm{Q}(\mathrm{x}) \mathrm{dx}+\int \frac{\mathrm{R}(\mathrm{x})}{\mathrm{ax}^{2}+\mathrm{bx}+\mathrm{c}} \mathrm{dx}$
$\Rightarrow \int \frac{x^{2}+1}{x^{2}-5 x+6} d x=\int\left(\frac{5 x-5}{x^{2}-5 x+6}+1\right) d x$
$=5 \int \frac{x-1}{x^{2}-5 x+6} d x+\int 1 d x$
Consider $\int \frac{x-1}{x^{2}-5 x+6} d x$
Let $x-1=\frac{1}{2}(2 x-5)+\frac{3}{2}$ and split,
$\Rightarrow \int\left(\frac{2 x-5}{2\left(x^{2}-5 x+6\right)}+\frac{3}{2\left(x^{2}-5 x+6\right)}\right) d x$
$\Rightarrow \frac{1}{2} \int \frac{2 x-5}{\left(x^{2}-5 x+6\right)} d x+\frac{3}{2} \int \frac{1}{x^{2}-5 x+6} d x$
Consider $\int \frac{2 x-5}{\left(x^{2}-5 x+6\right)} d x$
Let $u=x^{2}-5 x+6 \rightarrow d x=\frac{1}{2 x-5} d u$
$\Rightarrow \int \frac{2 x-5}{\left(x^{2}-5 x+6\right)} d x=\int \frac{2 x-5}{u} \frac{1}{2 x-5} d u$
$=\int \frac{1}{u} d u$
We know that $\int \frac{1}{x} d x=\log |x|+c$
$\Rightarrow \int \frac{1}{\mathrm{u}} \mathrm{du}=\log |\mathrm{u}|=\log \left|\mathrm{x}^{2}-5 \mathrm{x}+6\right|$
Now consider $\int \frac{1}{x^{2}-5 x+6} d x$
$\Rightarrow \int \frac{1}{x^{2}-5 x+6} d x=\int \frac{1}{(x-3)(x-2)} d x$
By partial fraction decomposition,
$\Rightarrow \frac{1}{(x-3)(x-2)}=\frac{A}{x-3}+\frac{B}{x-2}$
$\Rightarrow 1=A(x-2)+B(x-3)$
$\Rightarrow 1=A x-2 A+B x-3 B$
$\Rightarrow 1=(A+B) x-(2 A+3 B)$
$\Rightarrow A+B=0$ and $2 A+3 B=-1$
Solving the two equations,
$\Rightarrow 2 A+2 B=0$
$2 A+3 B=-1$
$-B=1$
$\therefore B=-1$ and $A=1$
$\Rightarrow \int \frac{1}{(x-3)(x-2)} d x=\int\left(\frac{1}{x-3}-\frac{1}{x-2}\right) d x$
$=\int \frac{1}{x-3} d x-\int \frac{1}{x-2} d x$
Consider $\int \frac{1}{x-3} d x$
Let $u=x-3 \rightarrow d x=d u$
$\Rightarrow \int \frac{1}{\mathrm{x}-3} \mathrm{dx}=\int \frac{1}{\mathrm{u}} \mathrm{du}$
We know that $\int \frac{1}{x} d x=\log |x|+c$
$\Rightarrow \int \frac{1}{u} d u=\log |u|=\log |x-3|$
Similarly $\int \frac{1}{x-2} d x$
Let $u=x-2 \rightarrow d x=d u$
$\Rightarrow \int \frac{1}{x-2} d x=\int \frac{1}{u} d u$
We know that $\int \frac{1}{x} d x=\log |x|+c$
$\Rightarrow \int \frac{1}{\mathrm{u}} \mathrm{du}=\log |\mathrm{u}|=\log |\mathrm{x}-2|$
Then,
$\Rightarrow \int \frac{1}{x^{2}-5 x+6} d x=\int \frac{1}{(x-3)(x-2)} d x=\int \frac{1}{x-3} d x-\int \frac{1}{x-2} d x$
$=\frac{1}{2}\left(\log \left|x^{2}-5 x+6\right|\right)+\frac{3}{2}(\log |x-3|-\log |x-2|)$
$=\frac{\log \left|x^{2}-5 x+6\right|}{2}+\frac{3 \log |x-3|}{2}-\frac{3 \log |x-2|}{2}$
Then,
$\Rightarrow \int \frac{x^{2}+1}{x^{2}-5 x+6} d x=5 \int \frac{x-1}{x^{2}-5 x+6} d x+\int 1 d x$
We know that $\int 1 \mathrm{dx}=\mathrm{x}+\mathrm{c}$
$\begin{aligned} \Rightarrow 5 \int \frac{x-1}{x^{2}-5 x+6} d x+\int 1 d x & \\=& \frac{5 \log \left|x^{2}-5 x+6\right|}{2}+\frac{15 \log |x-3|}{2}-\frac{15 \log |x-2|}{2}+x+c \end{aligned}$
$=\frac{5 \log |x-2| \log |x-3|}{2}+\frac{15 \log |x-3|}{2}-\frac{15 \log |x-2|}{2}+x+c$
$=x-5 \log |x-2|+10 \log |x-3|+c$
$\therefore \mathrm{I}=\int \frac{\mathrm{x}^{2}+1}{\mathrm{x}^{2}-5 \mathrm{x}+6} \mathrm{dx}=\mathrm{x}-5 \log |\mathrm{x}-2|+10 \log |\mathrm{x}-3|+\mathrm{c}$
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