({x+{12x-36}}+{x-{12x-36}}) dx=

Correct Answer is : {43(x−3)3/2+C,if x>623x+C,if 3≤x≤6.{cases}4/3(x-3)³/2}+C, & {if } x>6 \\ 2{3}x+C, & {if } 3≤ x≤6.{cases}{34(x−3)3/2+C,23x+C,if x>6if 3≤x≤6.

{cases}4/3(x-3)³/2}+C, & {if } x>6 \\ 2{3}x+C, & {if } 3≤ x≤6.{cases}

{cases}4/3(x-3)³/2}+C, & {if } 3≤ x≤6 \\ 2{3}x+C, & {if } x>6.{cases}

({x+{12x-36}}+{x-{12x-36}}) dx=

Correct Answer is : {43(x−3)3/2+C,if x>623x+C,if 3≤x≤6.{cases}4/3(x-3)³/2}+C, & {if } x>6 \\ 2{3}x+C, & {if } 3≤ x≤6.{cases}{34(x−3)3/2+C,23x+C,if x>6if 3≤x≤6.

Test Index

AP EAPCET 08-Jul-2022 Shift 2 Paper

Section: Mathematics

Question : 71 of 160

Marks: +1, -0

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

$2\sqrt{3}x+C,\qquad\forall x$
$\;\frac{4(x-3)^{3/2}}{2}+C,\forall x$
$\begin{cases}\frac{4}{3}(x-3)^{3/2}+C, & \text{if } x>6 \\ 2\sqrt{3}x+C, & \text{if } 3\le x\le6.\end{cases}$
$\begin{cases}\frac{4}{3}(x-3)^{3/2}+C, & \text{if } 3\le x\le6 \\ 2\sqrt{3}x+C, & \text{if } x>6.\end{cases}$

Validate

Solution: 👈: Video Solution

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

Put

t^2=12x-36

\Rightarrow x=\frac{t^2+36}{12}

and

2t\,dt=12\,dx \Rightarrow t\,dt=6\,dx

I=\int \left(\sqrt{\frac{t^2+36}{12}+t}+\sqrt{\frac{t^2+36}{12}-t}\right)\left(\frac{t}{6}\right)dt

I=\frac{1}{6}\int \frac{(t+6)+(t-6)}{\sqrt{12}}t\,dt

[\text{when } t^2>36]

=\frac{1}{6\sqrt{12}}\int (2t)t\,dt=\frac{1}{3\sqrt{12}}\int t^2\,dt

=\frac{1}{3\sqrt{12}}\left(\frac{t^3}{3}\right)+C=\frac{1}{9\sqrt{12}}(12x-36)^{\frac{3}{2}}+C

=\frac{(12)^{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

Now,

I=\frac{1}{6\sqrt{12}}\int (t+6-(t-6))t\,dt\quad[3\le x\le6]

=\frac{1}{6\sqrt{12}}\int 12t\,dt=\frac{\sqrt{12}}{6}\int t\,dt

=\frac{\sqrt{12}}{6}\left(\frac{t^2}{2}\right)+C'=\frac{1}{\sqrt{12}}(12x-36)+C'

=\sqrt{12}x+\left(\frac{-36}{\sqrt{12}}+C'\right)

\therefore I=2\sqrt{3}x+C,

3\le x\le6

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