Question

Find the value of the integral: ∫ (2x + 3)/((xy)(x^2 + 1)) dx

• A. \( \frac{1}{y} \ln(x^2 + 1) + C \)
• B. \( \frac{2}{y} \ln|x| + \frac{3}{y} \tan^{-1}(x) + C \)
• C. \( \frac{2}{y} \ln|x| + \frac{3}{y} \ln(x^2 + 1) + C \)
• D. \( \frac{1}{y} \ln|x(x^2 + 1)| + C \)

Hint:

Look for opportunities to split integrals and apply substitution or partial fractions when the integrand is a rational function involving polynomials.

Answer 1

The Correct Option is B

We are given: \[ \int \frac{2x + 3}{(xy)(x^2 + 1)} \, dx = \frac{1}{y} \int \frac{2x + 3}{x(x^2 + 1)} \, dx \] Now split the integral: \[ = \frac{1}{y} \left[ \int \frac{2x}{x(x^2 + 1)} \, dx + \int \frac{3}{x(x^2 + 1)} \, dx \right] \] The first term: \[ \int \frac{2x}{x(x^2 + 1)} dx = \int \frac{2}{x^2 + 1} dx = 2 \tan^{-1}(x) \] The second term: \[ \int \frac{3}{x(x^2 + 1)} dx = 3 \int \frac{1}{x(x^2 + 1)} dx \] This integral can be solved using partial fractions. The result is: \[ 3 \int \frac{1}{x(x^2 + 1)} dx = 3 \ln|x| - 3 \tan^{-1}(x) \] So, the combined result is: \[ \frac{1}{y} \left[ 2 \tan^{-1}(x) + 3 \ln|x| - 3 \tan^{-1}(x) \right] + C = \frac{2}{y} \ln|x| + \frac{3}{y} \tan^{-1}(x) + C \]