Question

If $e^x(1+x){\sec }^{2}x{e}^{x}dx=f\left(x\right)+$ constant, then $f\left(x\right)$ is equal to

• A. $\cos \left(x{e}^x\right)$
• B. $\sin \left(x{e}^x\right)$
• C. $2{\tan }^{-1}\left(x\right)$
• D. $\tan \left(x{e}^x\right)$

Answer 1

Answer: (D)

$\tan \left(x{e}^x\right)$

Given that,
$e^x(1+x)\cdot {\sec }^{2}x{e}^{x}dx=f\left(x\right)+$ constant
Put $x{e}^x=t$ in LHS
$\Rightarrow (e^x+x{e}^x)dx=dt$
$\therefore$ LHS $={\sec }^{2}tdt$
$=\tan t+$ constant
$\Rightarrow \tan \left(x{e}^x\right)+$ constant $=f\left(x\right)+$ constant
$\Rightarrow f\left(x\right)=\tan \left(x{e}^x\right)$