Question

Solve $\int \frac{e^x(1+x)}{{\cos }^2\left(e^{x}x\right)}dx$ 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. $\cot \left(e{x}^x\right)+C$

Answer 1

Answer: (B)

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

Consider the given integral.

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

Let $t=e^{x}x$

$\frac{dt}{dx}=e^{x}x+e^x$

$dt=e^x(x+1)dx$

Therefore,

$I=\int \frac{dt}{{\cos }^{2}t}$

$I=\int {\sec }^{2}tdt$

$I=\tan t+C$

On putting the value of $t$, we get

$I=\tan \left(e^{x}x\right)+C$

Hence, this is the answer.