Integrate $ \int \frac{1+x\cos(x)}{x(1-x^2(e^{2\sin(x)}))}dx $ $$ \int \frac{1+x\cos x}{x(1-x^2e^{2\sin x})}dx  $$
Attempt:
I substituted $(1-xe^{2\sin(x)})$ by $u$ and tried from there by differentiating it. But I get stuck midway.
 A: Just substitute,
   $$xe^{\sin(x)} = t$$
$$e^{\sin(x)}(1+x\cos(x))dx = dt $$
then your integral will become
$$ \int \frac{1}{t(1-t^2)}dt  $$
then it is 
$$ \int \frac{1}{t}+\frac{1}{2(1-t)}-\frac{1}{2(1+t)} dt  $$
$$=\ln(t)-\frac{\ln(1-t)}{2}-\frac{\ln(1+t)}{2}+c$$
then reverting the substitution
$$=\ln(xe^{\sin(x)} )-\frac{\ln(1-xe^{\sin(x)} )}{2}-\frac{\ln(1+xe^{\sin(x)} )}{2}+c$$
$$=-\ln{\sqrt{x^{-2}e^{-2\sin(x)}-1}}+c$$
A: HINT:
Note that we can write
$$\begin{align}
\frac{1+x\cos(x)}{x(1-x^2e^{2\sin(x)})}&=(1+x\cos(x))\left(\frac{1}{x(1-x^2e^{2\sin(x)})}\right)\\\\
&=(1+x\cos(x))\left(\frac{1}{x}+\frac{xe^{2\sin(x)}}{1-x^2e^{2\sin(x)}}\right)\\\\
&=\frac1x+\cos(x)+\frac{(x+x^2\cos(x))e^{2\sin(x)}}{1-x^2e^{2\sin(x)}} \tag 1
\end{align}$$
Can you find a substitution for the third term in $(1)$ that facilitates quick integration?
A: Substitute $1-x^2e^{2\sin x} = u$
This gives $$-[2xe^{2\sin x} + 2x^2 \cos x e^{2\sin x}] dx = du \\
-2xe^{2 \sin x} (1+x\cos x) dx = du$$
Putting the value of $dx$ in your integral, we get 
$$\int \frac{du}{-2x^2e^{2\sin x}(u)}$$
We know that $$\begin{align} 1-x^2 e^{2\sin x} &= u \\
x^2 e^{2\sin x} &= 1-u \\
-2x^2 e^{2\sin x} &= 2(u-1)\end{align}$$
This turns the integral to 
$$\int \frac{du}{2u(u+1)}$$
Use partial fractions to go further.
