int(sqrt{x+sqrt{12x-36}}+sqrt{x-sqrt{12x-36}} | vedclass

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Question

(C) Let $I = \int(\sqrt{x+\sqrt{12x-36}}+\sqrt{x-\sqrt{12x-36}}) dx$.The expression inside the square root is $x \pm \sqrt{12(x-3)}$.We can write $x \

Vedclass · Accepted Answer

$\begin{cases} \frac{4}{3}(x-3)^{3/2}+C, & 	ext{if } x > 6 \ 2\sqrt{3}x+C, & 	ext{if } 3 \leq x \leq 6 \end{cases}$

Vedclass · Answer

(C) Let $I = \int(\sqrt{x+\sqrt{12x-36}}+\sqrt{x-\sqrt{12x-36}}) dx$.The expression inside the square root is $x \pm \sqrt{12(x-3)}$.We can write $x \pm \sqrt{12(x-3)} = x \pm 2\sqrt{3(x-3)} = (x-3) \pm 2\sqrt{3(x-3)} + 3 = (\sqrt{x-3} \pm \sqrt{3})^2$.Thus,$\sqrt{x \pm \sqrt{12x-36}} = |\sqrt{x-3} \pm \sqrt{3}|$.The integral becomes $I = \int (\sqrt{x-3} + \sqrt{3} + |\sqrt{x-3} - \sqrt{3}|) dx$.Case $1$: If $x > 6$,then $\sqrt{x-3} > \sqrt{3}$,so $|\sqrt{x-3} - \sqrt{3}| = \sqrt{x-3} - \sqrt{3}$.$I = \int (\sqrt{x-3} + \sqrt{3} + \sqrt{x-3} - \sqrt{3}) dx = \int 2\sqrt{x-3} dx = 2 \cdot \frac{2}{3}(x-3)^{3/2} + C = \frac{4}{3}(x-3)^{3/2} + C$.Case $2$: If $3 \leq x \leq 6$,then $\sqrt{x-3} \leq \sqrt{3}$,so $|\sqrt{x-3} - \sqrt{3}| = \sqrt{3} - \sqrt{x-3}$.$I = \int (\sqrt{x-3} + \sqrt{3} + \sqrt{3} - \sqrt{x-3}) dx = \int 2\sqrt{3} dx = 2\sqrt{3}x + C$.Therefore,the solution is $\begin{cases} \frac{4}{3}(x-3)^{3/2}+C, & 	ext{if } x > 6 \ 2\sqrt{3}x+C, & 	ext{if } 3 \leq x \leq 6 \end{cases}$.

$\int(\sqrt{x+\sqrt{12x-36}}+\sqrt{x-\sqrt{12x-36}}) dx=$

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