Evaluate: `int(x^2+1)/(x^2-5x+6)dx`

Explanation

Since degree of numerator is not less than degree of denominator so this is not a proper rational function.
`:.(x^(2)+1)/(x^(2)-5x+6)=1+(5x-5)/(x^(2)-5x+6)`………`(i)`
Let `(5x-5)/(x^(2)-5x+6)=(5x-5)/((x-2)(x-3))`
`implies(5x-5)/((x-2)(x-3))=(A_(1))/((x-2))+(A_(2))/((x-3))`........`(ii)`
So that, `5x-5=A_(1)(x-3)+A_(2)(x-2)`
Now, equating coefficient of `x` and constant terms on both sides, we get,
`A_(1)+A_(2)=5` and `3A_(1)+2A_(2)=5`
`impliesA_(1)=-5`, `A_(2)=10`
Hence, `(5x-5)/((x-2)(x-3))=(-5)/(x-2)+(10)/(x-3)` [From equation `(ii)`]
`implies(x^(2)+1)/(x^(2)-5x+6)=1-(5)/(x-2)+(10)/(x-3)` [From equation `(i)`]
Integrating both sides,
`int(x^(2)+1)/(x^(2)-5x+6)dx=intdx-int(5)/(x-2)dx+int(10)/(x-3)dx`
`impliesint(x^(2)+1)/(x^(2)-5x+6)dx=x-5log|x-2|+10log|x-3|+C`