Question
Find $\int \frac{x^{2}+1}{x^{2}-5 x+6} d x$
✓
Answer
Here the integrand $\frac{x^{2}+1}{x^{2}-5 x+6}$ is not proper rational function, so we divide $x^2 + 1$ by $x^2 - 5x + 6$ and find that
$\frac{x^{2}+1}{x^{2}-5 x+6}=1+\frac{5 x-5}{x^{2}-5 x+6}=1+\frac{5 x-5}{(x-2)(x-3)}$
Let $\frac{5 x-5}{(x-2)(x-3)}=\frac{A}{x-2}+\frac{B}{x-3}$
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, $\frac{x^{2}+1}{x^{2}-5 x+6}=1-\frac{5}{x-2}+\frac{10}{x-3}$
Therefore, $\int \frac{x^{2}+1}{x^{2}-5 x+6} d x=\int d x-5 \int \frac{1}{x-2} d x+10 \int \frac{d x}{x-3}$
= x - 5 log | x - 2| + 10 log | x - 3| + C.
$\frac{x^{2}+1}{x^{2}-5 x+6}=1+\frac{5 x-5}{x^{2}-5 x+6}=1+\frac{5 x-5}{(x-2)(x-3)}$
Let $\frac{5 x-5}{(x-2)(x-3)}=\frac{A}{x-2}+\frac{B}{x-3}$
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, $\frac{x^{2}+1}{x^{2}-5 x+6}=1-\frac{5}{x-2}+\frac{10}{x-3}$
Therefore, $\int \frac{x^{2}+1}{x^{2}-5 x+6} d x=\int d x-5 \int \frac{1}{x-2} d x+10 \int \frac{d x}{x-3}$
= x - 5 log | x - 2| + 10 log | x - 3| + C.
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