The original integral was $$\int_{-\ln{2}}^{\ln{2}} \frac{(\sin{x} +x)^{\frac{1}{3}}}{e^x} \text{d}x$$ one can transform this with properties of definite integrals $(u = a + b - x)$ into $$\frac{1}{2} \int_{-\ln{2}}^{\ln{2}} \frac{(\sin{x}+x)^{\frac{1}{3}}}{e^x}-\frac{(\sin{x}+x)^{\frac{1}{3}}}{e^{-x}} \text{d}x$$ which is then able to be transformed into the much more approachable looking $$-2\int_{0}^{\ln{2}} (\sin{x}+x)^\frac{1}{3}\cdot \sinh{x} \text{d}x$$ but unfortunately I am unsure where to proceed from here. Help would be appreciated!
Edit: I've flagged my post as duplicate. See Evaluating the integral $\int_{-\ln 2}^{\ln 2} e^{-x}(\sin x+x)^{1/3}dx$. Thank you to John Omielan for the help!
If the question is well formulated, note that is trivial. You are integrating a continuous function between $a$ and $b$ with $a=\ln 2=b$. Since the two bounds are the same, the integral is zero.