Question

If $\displaystyle \int \frac{x^2+1}{(x-3)(x-2)}\,dx = Px + Q\log|x-3| + R\log|x-2| + c$, where $c$ is constant of integration, then the values of $P,Q,R$ are respectively

• A. $0,\,10,\,5$
• B. $0,\,10,\,-5$
• C. $1,\,10,\,5$
• D. $1,\,10,\,-5$

Hint:

Always simplify rational functions using division before applying partial fractions.

Answer 1

The Correct Option is D

Step 1: Performing algebraic division.
\[ \frac{x^2+1}{(x-3)(x-2)} = 1 + \frac{5x-5}{(x-3)(x-2)} \]
Step 2: Using partial fractions.
\[ \frac{5x-5}{(x-3)(x-2)} = \frac{A}{x-3} + \frac{B}{x-2} \] \[ 5x-5 = A(x-2) + B(x-3) \]
Step 3: Finding constants.
Putting $x=3$: \[ 15-5 = A(1) \Rightarrow A = 10 \] Putting $x=2$: \[ 10-5 = -B \Rightarrow B = -5 \]
Step 4: Integrating termwise.
\[ \int \frac{x^2+1}{(x-3)(x-2)}dx = \int \left[1 + \frac{10}{x-3} - \frac{5}{x-2}\right] dx \] \[ = x + 10\log|x-3| - 5\log|x-2| + c \]
Step 5: Conclusion.
\[ P = 1,\quad Q = 10,\quad R = -5 \]