Evaluating the integral $\int_{-\ln 2}^{\ln 2} e^{-x}(\sin x+x)^{1/3}dx$ As the title states, I'm having quite a lot of trouble with the integral:
$$\int_{-\ln 2}^{\ln 2}  \frac{(\sin x+x)^{1/3}}{e^x}dx$$
The problem is that my standard (admittedly sparse) repertoire of tricks seems to have no effect at all. I can't think of any clever substitution that would work, and it looks like a nasty integral all-round. 
I was hoping to apply some symmetry arguments (especially considering the limits of integration), but the exponential function is neither even nor odd, so I don't see how this would work. 
I'm afraid I have absolutely no idea how to approach this. 
 A: With that $(\sin(x)+x)^{1/3}$, it's very unlikely to have a closed form antiderivative, and I don't see any way to get help from a contour integration.  Taking some advantage of symmetry, you can express your integral as $-2 \int_0^{\ln 2} (\sin(x)+x)^{1/3} \sinh(x)\; dx$.  The numerical value is approximately $-.4758557412777$.  The Inverse Symbolic Calculator doesn't find anything for this.  So I would hazard to guess there is no closed form.  
A: Use a substitution of $e^{-x}=u$
$\rightarrow -x=\ln{u}$
$\rightarrow -e^{-x}=du$
Exponential gets cancelled in integral 
$-\int [\sin {(\ln {u})}+\ln {u}] du$
This gives (I am omitting unnecessary details)
$ u\cos{\ln{u}}+u.\ln{u}- \int u.(\frac{1}{u}) du$
$= u\cos{\ln{u}}+u.\ln{u}- u $
Replacing $u=e^{-x}$
$=\frac{1}{2}\cos{\ln{2}}+\frac{1}{2}(-\ln{2}-1)-[2\cos(\ln{2})+2(\ln{2}-1)]$
Dear, I have omitted some obvious details, and I guess you should be able to carry forward. Should you still have problems, you can inbox.
Someone tells me Ihave missed the $\frac{1}{3}$. 
So starting from 
$-\int [\sin {(\ln {u})}+\ln {u}]^\frac{1}{3} du$
It can be rewritten as
$\int[\ln{u}(\frac{\sin{(\ln{u})}}{\ln{u}}+1)]^{\frac{1}{3}}$
$\int(\ln{u})^\frac{1}{3}[(\frac{\sin{(\ln{u})}}{\ln{u}}+1)]^{\frac{1}{3}}$
If we consider $[(\frac{\sin{(\ln{u})}}{\ln{u}}+1)]^{\frac{1}{3}}$, by a Mac Laurin series, we find that this limit is not defined. It is divergent. Hence a product having this component will also be divergent. 
This integral cannot be calculated manually.
