Question

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

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

Answer 1

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

Let $x{e}^x=t$ then $e^x(1+x)dx=dt$
$\int \frac{e^x(1+x)dx}{{\cos }^2\left(e^{x}x\right)}$
$=\int {\sec }^{2}tdt$
$=\tan t+c$
$=\tan \left(e^{x}x\right)+c$