Question

\( \int \sqrt{x^2 + 3x} dx = \)}

• A. \( \sqrt{x^2 + 3x} + \log ....... \)
• B. \( \frac{2x+3}{4} \sqrt{x^2 + 3x} - \frac{9}{8} \log (x + \frac{3}{2} + \sqrt{x^2 + 3x}) + c \)
• C. \( x \sqrt{x^2 + 3x} + ....... \)
• D. \( x + 3\sqrt{x^2 + 3x} + ....... \)

Hint:

Standard formula: $\int \sqrt{x^2-a^2} = \frac{x}{2}\sqrt{x^2-a^2} - \frac{a^2}{2}\log|x+\sqrt{x^2-a^2}|$.

Answer 1

The Correct Option is B

Step 1: Complete the Square
$x^2 + 3x = (x + 3/2)^2 - (3/2)^2$.
Step 2: Standard Integral Form
$\int \sqrt{u^2 - a^2} du = \frac{u}{2}\sqrt{u^2-a^2} - \frac{a^2}{2}\log|u + \sqrt{u^2-a^2}|$.
Step 3: Substitution
$u = x + 3/2, a = 3/2 \implies a^2/2 = 9/8$.
Result $= \frac{x+3/2}{2} \sqrt{x^2+3x} - \frac{9}{8} \log|x+3/2 + \sqrt{x^2+3x}|$.
Step 4: Conclusion
Option (B) matches after simplifying the first term coefficient to $(2x+3)/4$.
Final Answer:(B)