Question 180-187 of 346: Integration and Its Applications (Section A: Common) | CUET (Common University Entrance Test) UG Mathematics Common (319) | Includes PYQs | Get Solutions with Detailed Explanations

${\displaystyle -\sqrt{2}}$ and ${\displaystyle \sqrt{2}\tan x}$

${\displaystyle \sqrt{2}}$ and ${\displaystyle \sqrt{2}\tan x}$

${\displaystyle \left[\frac{1}{2\sqrt{2}}\right]}$ and ${\displaystyle \sqrt{2}\tan x}$

${\displaystyle \left(\frac{2}{\sqrt{4}}\right)}$ and ${\displaystyle \left[\frac{1}{\sqrt{2}\cot x}\right]}$

${\displaystyle 2e^{\left(\left[\frac{(-x)}{2}\right]\right)}\sec \left(\frac{x}{2}\right)}$

${\displaystyle e^{\left(\left[\frac{(-x)}{2}\right]\right)}\sec \left(\frac{x}{2}\right)}$

${\displaystyle -e^{\left(\left[\frac{(-x)}{2}\right]\right)}\sec \left(\frac{x}{2}\right)}$

${\displaystyle -2e^{\left(\left[\frac{(-x)}{2}\right]\right)}\sec \left(\frac{x}{2}\right)}$

${\displaystyle \left[\frac{\left(13\right)}{7}\right]{\left[\frac{x+2}{x-5}\right]}^{\left(\left(\frac{1}{13}\right)\right)}}$

${\displaystyle \left[\frac{(-13)}{7}\right]{\left[\frac{x-5}{x+2}\right]}^{\left(\left(\frac{1}{13}\right)\right)}}$

${\displaystyle \left[\frac{\left(13\right)}{7}\right]{\left[\frac{x-5}{x+2}\right]}^{\left(\left(\frac{1}{13}\right)\right)}}$

${\displaystyle \left[\frac{(-13)}{7}\right]{\left[\frac{x+2}{x-5}\right]}^{\left(\left(\frac{1}{13}\right)\right)}}$

${\displaystyle (x-1)e^{\left([x+\left(\frac{1}{x}\right)]\right)}}$

${\displaystyle (x+1)e^{\left([x+\left(\frac{1}{x}\right)]\right)}}$

${\displaystyle x{e}^{\left([x+\left(\frac{1}{x}\right)]\right)}}$

${\displaystyle -x{e}^{\left([x+\left(\frac{1}{x}\right)]\right)}}$

1

${\displaystyle 2}$

${\displaystyle -1}$

${\displaystyle -2}$

${\displaystyle \log \left[\frac{(e^x+2012)}{(e^x+2013)}\right]}$

${\displaystyle \left[\frac{(e^x+2013)}{(e^x+2012)}\right]}$

${\displaystyle \left[\frac{(e^x+2012)}{(e^x+2013)}\right]}$

${\displaystyle \log \left[\frac{(e^x+2013)}{(e^x+2012)}\right]}$

${\displaystyle {\log }_{e}25}$

${\displaystyle {\log }_e\left(\frac{1}{5}\right)}$

${\displaystyle {\log }_e\left(\frac{1}{25}\right)}$

${\displaystyle \left[\frac{1}{{\log }_{e}5}\right]}$

${\displaystyle -x{\tan }^{-1}x}$

${\displaystyle x{\tan }^{-1}x}$

${\displaystyle 2x{\tan }^{-1}x}$

${\displaystyle -2x{\tan }^{-1}x}$