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: Let the integrand be simplified.
Let \[ I=\sqrt{x+\sqrt{12x-36}}+\sqrt{x-\sqrt{12x-36}} \] Squaring both sides, \[ I^2=x+\sqrt{12x-36}+x-\sqrt{12x-36} +2\sqrt{x^2-(12x-36)} \] \[ I^2=2x+2\sqrt{x^2-12x+36} \] \[ I^2=2x+2\sqrt{(x-6)^2} \] \[ I^2=2x+2|x-6| \]

Step 2: Consider the domain.
Since \[ \sqrt{12x-36} \] is defined, we need \[ 12x-36\geq 0 \] \[ x\geq 3 \] So, we consider two cases: \[ 3\leq x\leq 6 \] and \[ x>6 \]

Step 3: Case 1, when \(3\leq x\leq 6\).
Here, \[ |x-6|=6-x \] Therefore, \[ I^2=2x+2(6-x) \] \[ I^2=12 \] \[ I=2\sqrt{3} \] Hence, \[ \int I\,dx=\int 2\sqrt{3}\,dx \] \[ =2\sqrt{3}x+C \]

Step 4: Case 2, when \(x>6\).
Here, \[ |x-6|=x-6 \] Therefore, \[ I^2=2x+2(x-6) \] \[ I^2=4x-12 \] \[ I^2=4(x-3) \] \[ I=2\sqrt{x-3} \] Hence, \[ \int I\,dx=\int 2\sqrt{x-3}\,dx \] \[ =2\int (x-3)^{1/2}\,dx \] \[ =2\cdot \frac{(x-3)^{3/2}}{3/2}+C \] \[ =\frac{4}{3}(x-3)^{3/2}+C \]

Step 5: Final conclusion.
Therefore, \[ \boxed{ \begin{cases} \dfrac{4}{3}(x-3)^{3/2}+C, & x>6 2\sqrt{3}x+C, & 3\leq x\leq 6 \end{cases}} \]