Question

If ∫ (2x + 3)/((x - 1)(x^2 + 1)) dx = log_x {(x - 1)^(5/2)(x^2 + 1)^a} - (1/2) tan^(-1)x + C, then the value of a is:

• A. \( \frac{5}{4} \)
• B. \( -\frac{5}{3} \)
• C. \( -\frac{5}{6} \)
• D. \( -\frac{5}{4} \)

Hint:

Break rational expressions into partial fractions to simplify integration and match logarithmic terms to identify unknown constants.

Answer 1

The Correct Option is D

Given :
\[ \frac{2x + 3}{(x - 1)(x^2 + 1)} = \frac{A}{x - 1} + \frac{Bx + C}{x^2 + 1} \] Multiply both sides by \( (x - 1)(x^2 + 1) \):
\[ 2x + 3 = A(x^2 + 1) + (Bx + C)(x - 1) \] Expand the right side:
\[ A(x^2 + 1) + Bx(x - 1) + C(x - 1) = Ax^2 + A + Bx^2 - Bx + Cx - C \] Combine like terms:
\[ (A + B)x^2 + (-B + C)x + (A - C) \] Equate coefficients with \( 2x + 3 \):
\[ A + B = 0 \quad \text{(1)}
-B + C = 2 \quad \text{(2)}
A - C = 3 \quad \text{(3)} \] Solve the system of equations:
From (1): \( B = -A \)
Substitute into (2): \( -(-A) + C = 2 \Rightarrow A + C = 2 \quad \text{(4)} \)
From (3): \( A - C = 3 \)
Add (4) and (3):
\[ (A + C) + (A - C) = 2 + 3 \Rightarrow 2A = 5 \Rightarrow A = \frac{5}{2} \] Therefore, \( B = -\frac{5}{2} \), and \( C = 2 - A = 2 - \frac{5}{2} = -\frac{1}{2} \).
The partial fraction decomposition is:
\[ \frac{2x + 3}{(x - 1)(x^2 + 1)} = \frac{5/2}{x - 1} + \frac{-5x/2 - 1/2}{x^2 + 1} \] Integrate:
\[ \int \frac{5}{2(x - 1)} \, dx - \int \frac{5x/2 + 1/2}{x^2 + 1} \, dx \] Perform the integration:
\[ = \frac{5}{2} \ln|x - 1| - \frac{5}{4} \ln(x^2 + 1) - \frac{1}{2} \tan^{-1}x + C \] Compare with:
\[ \log_x \{(x - 1)^{\frac{5}{2}}(x^2 + 1)^a\} - \frac{1}{2} \tan^{-1}x + C \] Thus,
\[ a = -\frac{5}{4} \]