Q111 / 160 · AP EAMCET 2022 (08 Jul Shift 2) · Mathematics
$\int(\sqrt{x+\sqrt{12 x-36}}+\sqrt{x-\sqrt{12 x-36}}) d x=$
- A$2 \sqrt{3} x+C, \forall x$
- B$\frac{4(x-3)^{3 / 2}}{3}+C, \forall x$
- Correct answerC$\left\{\begin{array}{ccc}\frac{4}{3}(x-3)^{3 / 2}+C, & \text { if } & x>6 \\ 2 \sqrt{3} x+C & \text { if } & 3 \leq x \leq 6\end{array}\right.$
- D$\left\{\begin{array}{ccc}\frac{4}{3}(x-3)^{3 / 2}+C, & \text { if } & 3 \leq x \leq 6 \\ 2 \sqrt{3} x+C & \text { if } & x>6\end{array}\right.$
Solution
$I=\int(\sqrt{x+\sqrt{12 x-36}}+\sqrt{x-\sqrt{12 x-36}} d x$
Put $t^2=12 x-36$
$\Rightarrow x=\frac{t^2+36}{12}$
and $2 t d t=12 d x \Rightarrow t d t=6 d x$
$I=\int\left(\sqrt{\frac{t^2+36}{12}+t}+\sqrt{\frac{t^2+36}{12}-t}\right)\left(\frac{t}{6}\right) d t$
$I=\frac{1}{6} \int \frac{\{(t+6)+(t-6)\}}{\sqrt{12}} t d t \quad\left[\right.$ when $\left.t^2>36\right]$
$\begin{aligned} & =\frac{1}{6 \sqrt{12}} \int(2 t) t d t=\frac{1}{3 \sqrt{12}} \int t^2 d t \\ & =\frac{1}{3 \sqrt{12}}\left(\frac{t^3}{3}\right)+C=\frac{1}{9 \sqrt{12}}(12 x-36)^{\frac{3}{2}}+C \\ & =\frac{(1.2)^{3 / 2}}{9(12)^{1 / 2}}(x-3)^{3 / 2}+C=\frac{12}{9}(x-3)^{3 / 2}+C \\ & =\frac{4}{3}(x-3)^{3 / 2}+C, \text { if } x>6\end{aligned}$
Now, $I=\frac{1}{6 \sqrt{12}} \int\{t+6-(t-6)\} t d t[3 \leq x \leq 6]$
$\begin{aligned} & =\frac{1}{6 \sqrt{12}} \int 12 t d t=\frac{\sqrt{12}}{6} \int t d t \\ & =\frac{\sqrt{12}}{6}\left(\frac{t^2}{2}\right)+C^{\prime}=\frac{1}{\sqrt{12}}(12 x-36)+C^{\prime} \\ & =\sqrt{12} x+\left(\frac{-36}{\sqrt{12}}+C^{\prime}\right)\end{aligned}$
$\therefore \quad I=2 \sqrt{3} x+C$, if $3 \leq x \leq 6$
Hence, the correct answer is option: 'C'.