Question

\(\displaystyle \int \left(\sqrt{x+\sqrt{12x-36}}+\sqrt{x-\sqrt{12x-36}}\right)\,dx=\)

• A. \(2\sqrt{3}x+C,\ \forall x\)
• B. \(\dfrac{4(x-3)^{3/2}}{3}+C,\ \forall x\)
• C. \(\begin{cases} \dfrac{4}{3}(x-3)^{3/2}+C, & x>6 2\sqrt{3}x+C, & 3\leq x\leq 6 \end{cases}\)
• D. \(\begin{cases} \dfrac{4}{3}(x-3)^{3/2}+C, & 3\leq x\leq 6 2\sqrt{3}x+C, & x>6 \end{cases}\)

Hint:

When an expression contains \[ \sqrt{x+\sqrt{A}}+\sqrt{x-\sqrt{A}}, \] square the complete expression first. The term \[ \sqrt{x^2-A} \] often simplifies using modulus.

Answer 1

The Correct Option is C

Step 1: Square the expression to eliminate square roots.
Let S = √(x + √(12x–36)) + √(x – √(12x–36)). Squaring gives S² = 2x + 2√(x² – (12x–36)) = 2x + 2√((x–6)²) = 2x + 2|x–6|.

Step 2: Analyze domain and split cases.
Domain requires 12x – 36 ≥ 0 → x ≥ 3. For 3 ≤ x ≤ 6: |x–6| = 6–x → S² = 12 → S = 2√3. For x>6: |x–6| = x–6 → S² = 4x–12 = 4(x–3) → S = 2√(x–3).

Step 3: Integrate each case separately.
Case 1: ∫ 2√3 dx = 2√3 x + C. Case 2: ∫ 2√(x–3) dx = 2·(2/3)(x–3)^(3/2) + C = (4/3)(x–3)^(3/2) + C.

Step 4: Final Answer:
(4/3)(x–3)^(3/2) + C for x>6; 2√3 x + C for 3 ≤ x ≤ 6.