Find $\int \frac{x^2 + 1}{x^2 - 5x + 6} \, dx$

$x + 10 \ln|x - 3| - 5 \ln|x - 2| + C$

The correct answer is Option (1) → $x + 10 \ln|x - 3| - 5 \ln|x - 2| + C$

Here the integrand $\frac{x^2 + 1}{x^2 - 5x + 6}$ is not proper rational function, so we divide $x^2 + 1$ by $x^2 - 5x + 6$ and find that

$\frac{x^2+1}{x^2-5x+6} = 1 + \frac{5x-5}{x^2-5x+6} = 1 + \frac{5x-5}{(x-2)(x-3)}$

Let $\frac{5x-5}{(x-2)(x-3)} = \frac{A}{x-2} + \frac{B}{x-3}$

So that $5x - 5 = A(x - 3) + B(x - 2)$

Equating the coefficients of $x$ and constant terms on both sides, we get $A + B = 5$ and $3A + 2B = 5$. Solving these equations, we get $A = -5$ and $B = 10$.

Thus,

$\frac{x^2 + 1}{x^2 - 5x + 6} = 1 - \frac{5}{x-2} + \frac{10}{x-3}$

Therefore,

$\int \frac{x^2 + 1}{x^2 - 5x + 6} \, dx = \int dx - 5 \int \frac{1}{x-2} \, dx + 10 \int \frac{dx}{x-3}$

$= x - 5 \log |x-2| + 10 \log |x-3| + C$