The $\int{\left[ \dfrac{\left( 1+x \right){{e}^{x}}}{{{\sin }^{2}}\left( x{{e}^{x}} \right)} \right]dx}$ is
A. $-\cot \left( {{e}^{x}} \right)+c$
B. $\tan \left( x{{e}^{x}} \right)+c$
C. $\tan \left( {{e}^{x}} \right)+c$
D. $\cot \left( x{{e}^{x}} \right)+c$
E. $-\cot \left( x{{e}^{x}} \right)+c$

Hint: We first explain the terms $\dfrac{dy}{dx}$ where $y=f\left( x \right)$. We then need to integrate the equation once to find all the solutions of the integration. We take one arbitrary constant term for the integration. We use replacement of base value where $z=x{{e}^{x}}$ for $\int{\left[ \dfrac{\left( 1+x \right){{e}^{x}}}{{{\sin }^{2}}\left( x{{e}^{x}} \right)} \right]dx}$. We also use the integral theorem where $\int{{{\csc }^{2}}xdx}=-\cot x+c$.

Complete step by step answer:
We need to find the integral of $\int{\left[ \dfrac{\left( 1+x \right){{e}^{x}}}{{{\sin }^{2}}\left( x{{e}^{x}} \right)} \right]dx}$. We are going to change the base of the integral where we assume the new variable of $z=x{{e}^{x}}$. We take the new base and differentiate the equation $z=x{{e}^{x}}$. Differentiating both sides with respect to $x$, we get
\[\dfrac{d}{dx}\left( z \right)=\dfrac{d}{dx}\left( x{{e}^{x}} \right) \\
\Rightarrow \dfrac{dz}{dx}=x{{e}^{x}}+{{e}^{x}}=\left( 1+x \right){{e}^{x}} \]

Now we convert the differentiation into a differential form where \[dz=\left( 1+x \right){{e}^{x}}dx\].
We now replace all those values with \[dz=\left( 1+x \right){{e}^{x}}dx\] and $z=x{{e}^{x}}$ in the integration.
$\int{\left[ \dfrac{\left( 1+x \right){{e}^{x}}}{{{\sin }^{2}}\left( x{{e}^{x}} \right)} \right]dx}=\int{\dfrac{dz}{{{\sin }^{2}}z}} \\
\Rightarrow \int{\left[ \dfrac{\left( 1+x \right){{e}^{x}}}{{{\sin }^{2}}\left( x{{e}^{x}} \right)} \right]dx} =\int{{{\csc }^{2}}zdz} \\ $
Now we use the integral theorem of $\int{{{\csc }^{2}}xdx}=-\cot x+c$.
We get $\int{{{\csc }^{2}}zdz}=-\cot z+c$. We put the values where $z=x{{e}^{x}}$.
$\int{\left[ \dfrac{\left( 1+x \right){{e}^{x}}}{{{\sin }^{2}}\left( x{{e}^{x}} \right)} \right]dx}=-\cot z+c=-\cot \left( x{{e}^{x}} \right)+c$. Here $c$ is the integral constant.
The integral of $\int{\left[ \dfrac{\left( 1+x \right){{e}^{x}}}{{{\sin }^{2}}\left( x{{e}^{x}} \right)} \right]dx}$ is $-\cot \left( x{{e}^{x}} \right)+c$.

Hence, the correct option is E.

Note: We can also solve those integrations using the differential form directly. In that case the integral $\int{\left[ \dfrac{\left( 1+x \right){{e}^{x}}}{{{\sin }^{2}}\left( x{{e}^{x}} \right)} \right]dx}$ becomes $\int{{{\csc }^{2}}\left( x{{e}^{x}} \right)d\left( x{{e}^{x}} \right)}$. As the respective integral point is the same, we can use the formula $\int{{{\csc }^{2}}xdx}=-\cot x+c$.