Question

Evaluate \(\displaystyle\int\dfrac{x^2+1}{x^2-5x+6}\,dx\).

Hint:

Do polynomial long division first (equal degree top and bottom), then partial fractions on the remainder.

Answer 1

Step 1: Since numerator and denominator have equal degree, do polynomial division:
\[ \dfrac{x^2+1}{x^2-5x+6}=1+\dfrac{5x-5}{x^2-5x+6} \]

Step 2: Factor the denominator and set up partial fractions:
\(x^2-5x+6=(x-2)(x-3)\). Write \(\dfrac{5x-5}{(x-2)(x-3)}=\dfrac A{x-2}+\dfrac B{x-3}\).

Step 3: Solve for A and B:
\(5x-5=A(x-3)+B(x-2)\). At \(x=2\): \(5=A(-1)\Rightarrow A=-5\). At \(x=3\): \(10=B(1)\Rightarrow B=10\).

Step 4: Integrate term by term:
\[ \int\left[1-\dfrac5{x-2}+\dfrac{10}{x-3}\right]dx=x-5\ln|x-2|+10\ln|x-3|+C \]

Final Answer:
\[ \boxed{x-5\ln|x-2|+10\ln|x-3|+C} \]