Question

Evaluate the integral: \[ \int_0^{16} \frac{\sqrt{x}}{1 + \sqrt{x}} dx. \]

• A. \( 8 + 2\log 2 \)
• B. \( 8 + \log 2 \)
• C. \( 8 + 2\log 5 \)
• D. \( 8 + \log 5 \)

Hint:

Use \( t = \sqrt{x} \) substitution to simplify radicals. - Splitting fractions can make integration easier.

Answer 1

The Correct Option is C

Step 1: Use substitution
Let \( t = \sqrt{x} \), so that \( x = t^2 \) and \( dx = 2t dt \). Rewriting the integral: \[ I = \int_0^4 \frac{t}{1+t} \cdot 2t dt. \] \[ = 2 \int_0^4 \frac{t^2}{1+t} dt. \] Step 2: Split and integrate
\[ I = 2 \int_0^4 (t - 1 + \frac{1}{1+t}) dt. \] Solving each term: \[ I = 2 \left[ \frac{t^2}{2} - t + \log (1+t) \right] \Bigg|_0^4. \] Step 3: Compute the result
\[ I = 2 \left[ \left(\frac{16}{2} - 4 + \log 5 \right) - (0 - 0 + 0) \right]. \] \[ = 2 \times (8 - 4 + \log 5) = 8 + 2\log 5. \] Thus, the correct answer is \( \boxed{8 + 2\log 5} \).