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: Understanding the Question:
The question asks for the indefinite integral of a square root of a quadratic expression.
Step 2: Key Formula or Approach:
We use the method of completing the square to transform the quadratic into the form \( u^2 - a^2 \).
Then we apply the standard integral formula:
\[ \int \sqrt{u^2 - a^2} du = \frac{u}{2} \sqrt{u^2 - a^2} - \frac{a^2}{2} \log |u + \sqrt{u^2 - a^2}| + c \] Step 3: Detailed Explanation:
The expression is \( x^2 + 3x \).
Completing the square:
\[ x^2 + 3x = x^2 + 3x + \left(\frac{3}{2}\right)^2 - \left(\frac{3}{2}\right)^2 = \left(x + \frac{3}{2}\right)^2 - \frac{9}{4} \] Let \( u = x + \frac{3}{2} \) and \( a = \frac{3}{2} \). Then \( du = dx \).
The integral becomes:
\[ \int \sqrt{u^2 - a^2} du = \frac{u}{2} \sqrt{u^2 - a^2} - \frac{a^2}{2} \log |u + \sqrt{u^2 - a^2}| + c \] Substituting back \( u = x + \frac{3}{2} \), \( a^2 = \frac{9}{4} \), and \( u^2 - a^2 = x^2 + 3x \):
\[ = \frac{x + 3/2}{2} \sqrt{x^2 + 3x} - \frac{9/4}{2} \log |x + \frac{3}{2} + \sqrt{x^2 + 3x}| + c \] \[ = \frac{2x + 3}{4} \sqrt{x^2 + 3x} - \frac{9}{8} \log |x + \frac{3}{2} + \sqrt{x^2 + 3x}| + c \] Step 4: Final Answer:
The result is \( \frac{2x+3}{4} \sqrt{x^2 + 3x} - \frac{9}{8} \log (x + \frac{3}{2} + \sqrt{x^2 + 3x}) + c \).