$$
\begin{aligned}
& \text { Let } \mathrm{I}=\int \frac{(1+x \cos x)}{x\left[1-x^2\left(e^{\sin x}\right)^2\right]} \\
& d x=\int \frac{(1+x \cos x) e^{\sin x}}{x e^{\sin x}\left[1-\left(x e^{\sin x}\right)^2\right]} d x
\end{aligned}
$$

Let $x e^{\sin } x=t \Rightarrow e^{\sin x} \cdot(1+x \cos x) d x=d t$
$$
\begin{aligned}
& \text { Hence } I=\int \frac{d t}{t\left(1-t^2\right)}=\int\left[\frac{1}{t}-\frac{1}{2(t-1)}-\frac{1}{2(t+1)}\right] d t \\
& \Rightarrow \quad I=\ln |\mathrm{t}|-\frac{1}{2} \ln |(\mathrm{t}-1)|-\frac{1}{2} \ln |(\mathrm{t}+1)|+C \\
& =\frac{1}{2} \ln \left|\frac{\left(x e^{\sin x}\right)^2}{\left(x e^{\sin x}\right)^2-1}\right|+C
\end{aligned}
$$