Question:
$\int \frac{e^{x}(1+x)}{\cos ^{2}\left(e^{x} x\right)} d x$ 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. $\cot \left(e^{x}\right)+C$
Solution:
$\int \frac{e^{x}(1+x)}{\cos ^{2}\left(e^{x} x\right)} d x$
Let $e^{x} x=t$
$\Rightarrow\left(e^{x} \cdot x+e^{x} \cdot 1\right) d x=d t$
$e^{x}(x+1) d x=d t$
$\therefore \int \frac{e^{x}(1+x)}{\sec ^{2}\left(x^{x}\right)} d x=\int \frac{d t}{\cos ^{2} t}$
$=\int \sec ^{2} t d t$
$=\tan t+\mathrm{C}$
$=\tan \left(e^{x} \cdot x\right)+\mathrm{C}$
Hence, the correct answer is B.