Find the indefinite integral.
${\int }_{}^{}\frac{3-y^2}{2y}dy$

• A. $\frac{3}{2}\ln \mid y\mid -\frac{1}{4}{y}^2+C$
• B. $\frac{3}{2}+\frac{1}{4y^2}+C$
• C. $\frac{3}{2}-\frac{1}{4y^2}+C$
• D. $\frac{3}{2}\ln \mid y\mid -\frac{1}{4y^2}+C$

Answer: (A)

Verified step by step guidance

1. Rewrite the integrand \( \frac{3 - y^2}{2y} \) by splitting it into two separate terms: \( \frac{3}{2y} - \frac{y^2}{2y} \). Simplify each term to get \( \frac{3}{2y} - \frac{y}{2} \).
2. Set up the integral as \( \int \left( \frac{3}{2y} - \frac{y}{2} \right) \, dy \). Now, split the integral into two parts: \( \int \frac{3}{2y} \, dy - \int \frac{y}{2} \, dy \).
3. For the first term, \( \int \frac{3}{2y} \, dy \), factor out the constant \( \frac{3}{2} \) to get \( \frac{3}{2} \int \frac{1}{y} \, dy \). The integral of \( \frac{1}{y} \) is \( \ln|y| \), so this becomes \( \frac{3}{2} \ln|y| \).
4. For the second term, \( \int \frac{y}{2} \, dy \), factor out the constant \( \frac{1}{2} \) to get \( \frac{1}{2} \int y \, dy \). The integral of \( y \) is \( \frac{y^2}{2} \), so this becomes \( \frac{1}{2} \cdot \frac{y^2}{2} = \frac{y^2}{4} \).
5. Combine the results of the two integrals: \( \frac{3}{2} \ln|y| - \frac{y^2}{4} + C \), where \( C \) is the constant of integration.