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galactus · Mar 1, 2008, 05:35 AM

Here's another problem if anyone is interested.

Can you perform the integration? I am curious to see if anyone has a cool method.

Anyone who knows integration knows that $e^{\frac{x^{2}}{2}}$ is not easily integrated.

$\int{(x^{2}+1)e^{\frac{x^{2}}{2}}}dx$

galactus · Mar 10, 2008, 04:41 PM

It appears no one was interested in trying the problem. Not that anyone cares, but I will post my solution.

Parts or substitution do not work too well. Here is a rather obscure method that works well

when you have something of the form $p(x)e^{x}$

This is known as integration by recognition.

$\frac{d}{dx}[p(x)e^{\frac{x^{2}}{2}}]=(x^{2}+1)e^{\frac{x^{2}}{2}}$

Let $p(x)=ax^{2}+bx+c$

Then $\frac{d}{dx}(ax^{2}+bx+x)e^{\frac{x^{2}}{2}}$

By the product rule:

$(2ax+b)(e^{\frac{x^{2}}{2}})-(ax^{2}+bx+c)xe^{\frac{x^{2}}{2}}=<br /> <br /> (x^{2}+1)e^{\frac{x^{2}}{2}}$

Factor, group like terms and equate coefficients, we get:

$a=0, \;\ b=1, \;\ 2a+c=0$

$\frac{d}{dx}(xe^{\frac{x^{2}}{2}})=(x^{2}+1)\cdot{ e^{\frac{x^{2}}{2}}}$

Now integrate both sides and we get:

$\fbox{xe^{\frac{x^{2}}{2}}=\int{(x^{2}+1)e^{\frac{ x^{2}}{2}}dx}$

Capuchin · Mar 10, 2008, 05:24 PM

Hi gal, just wanted to say that I enjoy these, don't get disheartened that nobody attempted it.

Also, clean out your inbox, I might need to message you in the future :)

galactus · Mar 10, 2008, 05:32 PM

It's cleaned. It wasn't full. About half.

Capuchin · Mar 10, 2008, 05:43 PM

Last time I wanted to send you a message it was full! :o