Question

$\int \frac{e^x(1+x)}{{\cos }^2\left(e^{x}x\right)}dx$ equals

• A. $-cot\left(e{x}^x\right)+C$
• B. $tan\left(x{e}^x\right)+C$
• C. $tan\left(e^x\right)+C$
• D. $cot\left(e^x\right)+C$

Answer 1

The correct option is B $tan\left(x{e}^x\right)+C$

$\int \frac{e^x(1+x)}{{\cos }^2\left(e^{x}x\right)}dx$
Let $e^{x}x=t\Rightarrow (e^x\cdot x+e^x\cdot 1)dx=dt$
$\Rightarrow e^x(x+1)dx=dt$
$\therefore \int \frac{e^x(1+x)}{{\cos }^2\left(e^{x}x\right)}dx=\int \frac{dt}{{\cos }^{2}t}$
$=\int {\sec }^{2}tdt=\tan t+C=\tan (e^x\cdot x)+C$
Hence, the correct Answer is B.