$e^{x^2/2}\int e^{-x^2/2}(-x^3+x)\ dx$ turns out to be equal to $e^{x^2/2}[e^{-x^2/2}(x^2+1)] $
Is there a easier method of integrating such functions? I can't grasp how text book was able to integrate it so easily. They dont show the steps, but rather go straight to integrated function. How would you tackle such integration? Integration by parts of these fuctions proven to be tedious and time consuming as they require another integration by parts to be performed.
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$\begingroup$Hopefully you recognize $e^{-x^2}$ as a function which can not be integrated using regular functions. As such using 'by parts' as your first step with your parts being the exponential and the polynomial will be unsuccessful. Note that the derivative of $x^2$ contains $x$ and the second part of your integral has this as a factor so:
$$\int e^{-x^2/2}(-x^3+x)\ dx=\int e^{-x^2/2}(-x^2+1)x\ dx$$
Let $u=\frac{x^2}{2}$ so $du=x\ dx$.
$$=\int e^{-u}(-2u+1)du$$
$$=\int e^{-u}\ du-2\int u e^{-u}\ du$$
Then use 'by parts' on the second integral
$$=-e^{-u} - 2\left(-u e^u-\int -e^u\ du\right)$$
$$=-e^{-u} +2u e^{-u}+2e^{-u} +c$$
$$=2u e^{-u}+e^{-u} + c$$
$$=x^2e^{x^2/2}+e^{-x^2/2}+c$$
$$=e^{x^2/2}(x^2+1)+c$$
$\endgroup$ 1 $\begingroup$If, as in this case, the terms of the polynomial are all odd, the substitution $u = x^2$ works.
$$ J = \int e^{-x^2/2} (-x^3 + x)\; dx = \int e^{-u/2} \dfrac{-u+1}{2}\; du$$
Next, using the method of undetermined coefficients, a form for the antiderivative should be
$$ J = e^{-u/2} (a u + b) + C$$
Taking the derivative, we need $$ \dfrac{dJ}{du} = e^{-u/2} \left(-\frac{a u + b}{2} + a\right) = e^{-u/2} \dfrac{-u+1}{2}$$ Thus $a = 1$ and $a-b/2 = 1/2$, so $b = 1$. The result is
$$ J = e^{-u/2} \left(u + 1 \right) + C= e^{-x^2/2} \left( x^2 + 1\right) + C$$
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