$\int_{-1}^1 e^{-x^{4}}\cdot(1+\ln(x+\sqrt{x^2+1})+5x^3-4x^4)dx$ Is it possible to evaluate this integral $$\int_{-1}^1 e^{-x^{4}}\cdot(1+\ln(x+\sqrt{x^2+1})+5x^3-4x^4)dx$$
I got this question on my math quiz today, Solution not given yet, But is it even possible to evaulate this integral
 A: You can neglect the $\ln{}$ and the $x^3$ terms, as they are odd over a symmetric interval.  Thus we have
$$2 \int_0^1 dx \, e^{-x^4} (1-4 x^4) $$
Integrate by parts to get
$$4 \int_0^1 dx \, x^4 e^{-x^4} = - \int_0^1 d(e^{- x^4}) x =  [x e^{-x^4}]_1^0 + \int_0^1 dx \, e^{-x^4}$$
Thus the integral is
$$2 \int_0^1 dx \, e^{-x^4} - \frac{2}{e} - 2 \int_0^1 dx \, e^{-x^4} = \frac{2}{e} $$
A: Notice that $x\mapsto\ln (x+\sqrt{x^2+1})$ is an odd function since
\begin{align}
\ln [(-x)+\sqrt{(-x)^2+1}]&=\ln\left[\frac{\sqrt{x^2+1}-x}{1}\cdot\frac{\sqrt{x^2+1}+x}{\sqrt{x^2+1}+x}\right]\\
&=\ln\left[\frac{1}{\sqrt{x^2+1}+x}\right]\\
&=-\ln (x+\sqrt{x^2+1})
\end{align}
Also $x\mapsto x^3$ is odd, then we have
\begin{align}
\int_{-1}^1 e^{-x^{4}}\cdot(\ln(x+\sqrt{x^2+1})+5x^3)dx&=0
\end{align}
Since the integrand is an odd function. Then the integral becomes
\begin{align}
\int_{-1}^1 e^{-x^{4}}\cdot(1-4x^4)dx&=\int_{-1}^1e^{-x^4}dx-\int_{-1}^14x^4e^{-x^4}dx\\
&=\int_{-1}^1e^{-x^4}dx+\int_{-1}^1x (-4x^3e^{-x^4})dx\\
&=\int_{-1}^1e^{-x^4}dx+\left.xe^{-x^4}\right|_{-1}^1-\int_{-1}^1e^{-x^4}dx\\
&=e^{-1}-(-1)e^{-1}\\
&=\frac{2}{e}
\end{align}
Where integration by parts was used.
