Question

If $\int \frac{2x+3}{(x-1)(x^2+1)} dx = \log_e \left\{ (x - 1)^{\frac{5}{2}} (x^2 + 1)^a \right\} - \frac{1}{2} \tan^{-1} x + A$ where A is an arbitrary constant, then the value of $a$ is

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

Hint:

In partial fractions, if the power of $x$ in the numerator is lower than the denominator, the sum of your $x^2$ coefficients will always be zero.

Answer 1

The Correct Option is B

Step 1: Concept
Use partial fractions to decompose the integrand: $\frac{2x+3}{(x-1)(x^2+1)} = \frac{P}{x-1} + \frac{Qx+R}{x^2+1}$.

Step 2: Meaning
Find the constants: $P(x^2+1) + (Qx+R)(x-1) = 2x+3$. Let $x=1: 2P = 5 \implies P = 5/2$. Compare $x^2$ coeffs: $P+Q=0 \implies Q = -5/2$. Compare constants: $P-R=3 \implies R = 5/2 - 3 = -1/2$.

Step 3: Analysis
$\int \frac{5/2}{x-1} dx + \int \frac{-5/2x - 1/2}{x^2+1} dx$. $= \frac{5}{2} \log(x-1) - \frac{5}{4} \log(x^2+1) - \frac{1}{2} \tan^{-1} x$. Combine logs: $\log(x-1)^{5/2} + \log(x^2+1)^{-5/4}$.

Step 4: Conclusion
Comparing with the given form $(x^2+1)^a$, we find $a = -5/4$. Final Answer: (B)