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$$\int_{-1}^1 e^{-x^{4}}\cdot(1+\ln(x+\sqrt{x^2+1})+5x^3-4x^4)dx$$

I got this question on my math quiz today, Solution not given yet, But is it even possible to evaulate this integral

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2 Answers 2

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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.

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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

$$2 \int_0^1 dx \, e^{-x^4} - \frac{2}{e} - 2 \int_0^1 dx \, e^{-x^4} = \frac{2}{e} $$

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