Question

Evaluate \( \int \frac{3x+2}{(x^2+1)(x-2)}dx \).

Hint:

For partial fractions with \( x^2+1 \), remember that the second term \( \frac{Bx+C}{x^2+1} \) often splits into a logarithmic term (from \( Bx \)) and a \( \tan^{-1} \) term (from \( C \)).

Answer 1

Step 1: Understanding the Concept:
To integrate a rational function with a quadratic factor in the denominator, we use partial fraction decomposition.
Step 2: Detailed Explanation:
Let \( \frac{3x+2}{(x^2+1)(x-2)} = \frac{A}{x-2} + \frac{Bx+C}{x^2+1} \).
Multiply by the denominator: \( 3x+2 = A(x^2+1) + (Bx+C)(x-2) \).
Put \( x = 2 \): \( 3(2)+2 = A(2^2+1) \implies 8 = 5A \implies A = \frac{8}{5} \).
Equate \( x^2 \) coefficients: \( 0 = A + B \implies B = -A = -\frac{8}{5} \).
Equate constants: \( 2 = A - 2C \implies 2 = \frac{8}{5} - 2C \implies 2C = \frac{8}{5} - 2 = -\frac{2}{5} \implies C = -\frac{1}{5} \).
Wait, recalculating constants...
Equate \( x \) coefficients: \( 3 = C - 2B \implies 3 = C - 2(-\frac{8}{5}) \implies 3 = C + \frac{16}{5} \implies C = 3 - \frac{16}{5} = -\frac{1}{5} \).
Correct constants are \( A = 8/5, B = -8/5, C = -1/5 \).
Now integrate:
\[ I = \int \frac{8/5}{x-2} dx + \int \frac{-8/5 x - 1/5}{x^2+1} dx \]
\[ I = \frac{8}{5} \log|x-2| - \frac{4}{5} \int \frac{2x}{x^2+1} dx - \frac{1}{5} \int \frac{1}{x^2+1} dx \]
\[ I = \frac{8}{5} \log|x-2| - \frac{4}{5} \log(x^2+1) - \frac{1}{5} \tan^{-1}x + C \]
Note: The provided options might vary based on the specific algebraic calculation steps of the question sheet creator.
Step 3: Final Answer:
The result is \( \frac{8}{5}\log|x-2| - \frac{4}{5}\log(x^2+1) - \frac{1}{5}\tan^{-1}x + C \).