What is the integral of ∫ e^ (x^2 +sinx) dx ?

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What is the integral of ∫ e^ (x^2 +sinx) dx


I got the answer e^(x^2+sinx)/(2x+cosx) but I know that is wrong. I don't understand how you treat the + part for an e^() problem.
 
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There is presumably no closed form solution involving elementary functions.

Mathematica agrees: http://integrals.wolfram.com/index.jsp?expr=e^(x^2+%2B+sin(x))&random=false
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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