$e^{x^2/2}\int e^{-x^2/2}(-x^3+x)\ dx$ 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.

Sorry for the link, it said I need 10 rep points to post the photo

  • $\begingroup$ I don't understand what you're asking. If you have a product of functions of a single variable, you pretty much need to integrate by parts unless one of the functions is the derivative of the other. $\endgroup$ – ÍgjøgnumMeg Jun 8 '16 at 1:32
  • $\begingroup$ for such product of functions, only method possible is integration by parts? $\endgroup$ – Sysnaptic Jun 8 '16 at 1:36
  • $\begingroup$ Look up "Reduction Formulae". With products of trigonometric functions and exponentials, you'll end up in an endless loop. $\endgroup$ – ÍgjøgnumMeg Jun 8 '16 at 1:41
  • $\begingroup$ You'll need to post in Latex, otherwise your post is likely to get down voted and/or closed. See here for formatting help. $\endgroup$ – mattos Jun 8 '16 at 1:44

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

$$=-e^{-u} - 2\left(-u e^u-\int -e^u\ du\right)$$

$$=-e^{-u} +2u e^{-u}+2e^{-u} +c$$

$$=2u e^{-u}+e^{-u} + c$$



  • $\begingroup$ Wow! Thank you so much for your answer. This answers everything I was curious about $\endgroup$ – Sysnaptic Jun 8 '16 at 2:55

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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