# antiderivative of e^(2x+x^2)

• May 3rd 2009, 03:31 PM
electricsparks
antiderivative of e^(2x+x^2)
How come the antiderivative of e^(2x+x^2) isn't e^(2x+x^2)/(x+2x)??
I know the anti derivative of e^(x^2) cannot be defined using elem. functions, so i guess e^(2x+x^2) is kinda like that, but it seemed to make sense to me when I antiderived it using substitution like this :

S e^u = e^u + C

let u= 2x+x^2
du = 2+2x dx
dx= du/2+2x

Then:
S e^(2x+x^2) dx= S e^u * du/2+2x = e^(2x+x^2)/2+2x

Doesn't that make sense? what am I doing wrong? or is that the antiderivative of e^(2x+x^2) altho I'm pretty sure it isn't?? does this have something to do with chain rule??

thanks!
• May 3rd 2009, 03:43 PM
HallsofIvy
Quote:

Originally Posted by electricsparks
How come the antiderivative of e^(2x+x^2) isn't e^(2x+x^2)/(x+2x)??
I know the anti derivative of e^(x^2) cannot be defined using elem. functions, so i guess e^(2x+x^2) is kinda like that, but it seemed to make sense to me when I antiderived it using substitution like this :

S e^u = e^u + C

let u= 2x+x^2
du = 2+2x dx
dx= du/2+2x

Then:
S e^(2x+x^2) dx= S e^u * du/2+2x = e^(2x+x^2)/2+2x

Doesn't that make sense? what am I doing wrong? or is that the antiderivative of e^(2x+x^2) altho I'm pretty sure it isn't?? does this have something to do with chain rule??

thanks!

What you have done is not right because you cannot have "x" in the "du" integral- since x is not a constant, you cannot treat that remaining "2+ 2x" as a constant. It's not so much a matter of the "chain rule" as it is the "quotient rule". If you were to differentiate $e^{2x+x^2}/(2+2x)$ you cannot just differentiate the " $e^{2x+x^2}$" and cancel the "2+2x" in the denominator. $\left(e^{2x+x^2}/(2+2x)\right)'= \frac{[e^{2x+ x^2}]'(2+2x)- e^{2x+x^2}(2+2x)'}{(2+2x)^2}$. That, of course, is NOT $e^{2x+ x^2}$

What you can do is "complete the square": write $2x+ x^2$ as $2x+ x^2+ 1- 1$= $(x+1)^2- 1$ so that $e^{2x+ x^2}= e^{(x+1)^2- 1}= e^{-1}e^{(x+1)^2}$. Now, $\int e^{2x+ x^2}dx= e^{-1}\int e^{(x+1)^2} dx$ and you can use the linear substitution u= x+ 1 so that du= dx. Now the integral is $e^{-1}\int e^{u^2}du$. Of course that integral cannot be done in any simple form, but now you can write it in terms of the "error function" Erf(x).
• May 3rd 2009, 06:06 PM
Sine
well, lets try using integration by parts here...

we have,

http://latex.codecogs.com/gif.latex?...x}e^{x^{2}})dx

Lets take the function
http://latex.codecogs.com/gif.latex?\large%20e^{2x}

as the second(as we can integrate it easily)

& use by parts. try to evaluate it now.

Spoiler:

http://latex.codecogs.com/gif.latex?...2x}e^{x^{2}}dx

which again on applying by parts in the second part
(taking http://latex.codecogs.com/gif.latex?\large%20e^{2x} as the first function now)

http://latex.codecogs.com/gif.latex?...e^{x^{2}}dx)dx

both the parts can now be integrated as the last part can again be written as a factor of the original integrand if you know what i mean.
• May 3rd 2009, 08:30 PM
mr fantastic
Quote:

Originally Posted by Sine
well, lets try using integration by parts here...

we have,

http://latex.codecogs.com/gif.latex?...x}e^{x^{2}})dx

Lets take the function
http://latex.codecogs.com/gif.latex?\large%20e^{2x}

as the second(as we can integrate it easily)

& use by parts. try to evaluate it now.

Spoiler:

http://latex.codecogs.com/gif.latex?...2x}e^{x^{2}}dx

which again on applying by parts in the second part
(taking http://latex.codecogs.com/gif.latex?\large%20e^{2x} as the first function now)

http://latex.codecogs.com/gif.latex?...e^{x^{2}}dx)dx

both the parts can now be integrated as the last part can again be written as a factor of the original integrand if you know what i mean.

I can say this with total confidence because, as earlier implied, the integral does not have an answer that is expressible using a finite number of elementary functions. This is a mathematical fact (and can be proved), it is not an intellectual shortcoming on the part of myself or anyone else.
• May 3rd 2009, 09:19 PM
Sine
Quote:

Originally Posted by mr fantastic

I can say this with total confidence because, as earlier implied, the integral does not have an answer that is expressible using a finite number of elementary functions. This is a mathematical fact (and can be proved), it is not an intellectual shortcoming on the part of myself or anyone else.

well i just used whatever has been taught to me & integrated the function.

so does a calc student need to know this as a fact?

& yes, the method gives me http://latex.codecogs.com/gif.latex?...{x^{2}}e^{2x}) as the answer which definitely does not yield the same expression on differentiating it.
• May 4th 2009, 05:57 AM
HallsofIvy
Quote:

Originally Posted by Sine
well i just used whatever has been taught to me & integrated the function.

Unfortunately, you used it incorrectly.

Quote:

so does a calc student need to know this as a fact?
Know what?

Quote:

& yes, the method gives me http://latex.codecogs.com/gif.latex?...{x^{2}}e^{2x}) as the answer which definitely does not yield the same expression on differentiating it.
and therefore it is wrong.

The problem is that when you do that final integration that you just say "can again be written as a factor of the original integrand" everything cancels and you just get $\int e^x e^{x^2} dx$ again.
• May 4th 2009, 07:13 AM
Sine
• May 4th 2009, 07:17 AM
TheEmptySet
As Mr. Fantastic said check your work!!

$\frac{dI}{dx}=\frac{1}{2}\left( 2e^{2x}e^{x^2}+e^{2x}e^{x^2}(2x)=(1+x){e^{2x+x^2}} \right)$

THis does NOT equal the original.
• May 4th 2009, 07:42 AM
Sine
Quote:

Originally Posted by TheEmptySet
As Mr. Fantastic said check your work!!

$\frac{dI}{dx}=\frac{1}{2}\left( 2e^{2x}e^{x^2}+e^{2x}e^{x^2}(2x)=(1+x){e^{2x+x^2}} \right)$

THis does NOT equal the original.

yes. the answer i got is the integral of (x+1)e^(2x)e^(x^2))

so do you identify functions like these? which you cannot integrate?
• May 4th 2009, 07:43 AM
Sine
Quote:

Originally Posted by Sine
yes. the answer i got is the integral of (x+1)e^(2x)e^(x^2))

so do you identify functions like these? which you cannot integrate?

i still cant find my mistake in the working
• May 4th 2009, 08:16 AM
TheEmptySet
@ Sine

$\int e^{2x}e^{x^2}dx$

Let $u=e^{x^2} \implies du=2xe^{x^2}dx$

$dv=e^{2x}dx \implies v=\frac{1}{2}e^{2x}$

$\int e^{2x}e^{x^2}dx=\frac{1}{2}e^{2x}e^{x^2}-\int xe^{2x}e^{x^2}dx$

So I think this is what you mean if we let

$u=e^{2x} \implies du=2e^{2x}dx$ and

$dv=xe^{x^2}dx \implies v=\frac{1}{2}e^{x^2}$

So putting this in above you get

$\int e^{2x}e^{x^2}dx=\frac{1}{2}e^{2x}e^{x^2}-\left(\frac{1}{2}e^{2x}e^{x^2}-\int e^{2x}e^{x^2}dx \right)$

This is just the identity

$\int e^{2x}e^{x^2}dx=\int e^{2x}e^{x^2}dx$
• May 4th 2009, 09:55 AM
Sine
Quote:

Originally Posted by TheEmptySet
@ Sine

$\int e^{2x}e^{x^2}dx$

Let $u=e^{x^2} \implies du=2xe^{x^2}dx$

$dv=e^{2x}dx \implies v=\frac{1}{2}e^{2x}$

$\int e^{2x}e^{x^2}dx=\frac{1}{2}e^{2x}e^{x^2}-\int xe^{2x}e^{x^2}dx$

So I think this is what you mean if we let

$u=e^{2x} \implies du=2e^{2x}dx$ and

$dv=xe^{x^2}dx \implies v=\frac{1}{2}e^{x^2}$

So putting this in above you get

$\int e^{2x}e^{x^2}dx=\frac{1}{2}e^{2x}e^{x^2}-\left(\frac{1}{2}e^{2x}e^{x^2}-\int e^{2x}e^{x^2}dx \right)$

This is just the identity

$\int e^{2x}e^{x^2}dx=\int e^{2x}e^{x^2}dx$

thanks . i got it. but how do we identify such functions? do they have something in common?