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Yeah I did try searching how to integrate $e^{x^2}$ and mostly I stumbled upon how a similar but not this function called Gaussian function $e^{-x^2}$ is un-integrable , now I was given to solve a differential equation $$xe^{x^2}dx + (y^5-1)dy = 0$$ so how do I proceed with the part of integrating $xe^{x^2}dx$ ? I have tried a lot, any help will be appreciated.

Stella Biderman
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Arnav Das
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  • So basically you want to know how to find $\int xe^{x^2}dx$? – Aditya Agarwal Dec 08 '15 at 07:43
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    Just if You are interested in integrating $e^{^-x^2}$, using Fubini and polar coordinates gives $(\int_{-\infty}^{\infty}e^{-x^2}dx)^2=\int_{-\infty}^{\infty}e^{-x^2}dx \int_{\infty}^{\infty}e^{-y^2}dy=\int_{\mathbb{R}^2}e^{-(x^2+y^2)}d\lambda_2= \int_0^{\infty}2\pi r e^{-r^2}dr=\pi$. – Peter Melech Dec 08 '15 at 07:57

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To find $$\int xe^{x^2}dx$$ Set $t=x^2$, $$\frac12\int e^tdt=\frac12e^{x^2}+C$$ Beware, changing the integrand from $xe^{x^2} $ to $e^{x^2}$ makes a hell lot of a difference. Here, you have a $f(g(x))g'(x)$ form, but you changed it to $f(g(x))$ form.

Aditya Agarwal
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