Integration of exp(x^2)*(x^2-1)/x^3

  • Thread starter notnottrue
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  • #1
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Homework Statement


I need to find the integral of [itex]\frac{exp(x^{2})(x^{2}-1)}{x^{3}}[/itex]

Homework Equations


Wolfram returns this answer
[itex]\frac{exp(x^{2})}{2x}[/itex]


The Attempt at a Solution


I orginally had the integral in the form of:
[itex]\frac{exp(x^{2})(x^{2})}{x^{3}}[/itex]-[itex]\frac{exp(x^{2})}{x^{3}}[/itex]

But I dont know how to integrate either!

Any help or suggestions will be appreciated!
 

Answers and Replies

  • #2
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Well, it's not necessary, but you can let u=x2. That should make it a little easier. You should use the chain, product and quotient rules.
 
  • #3
SammyS
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Homework Statement


I need to find the integral of [itex]\frac{exp(x^{2})(x^{2}-1)}{x^{3}}[/itex]

Homework Equations


Wolfram returns this answer
[itex]\frac{exp(x^{2})}{2x}[/itex]

The Attempt at a Solution


I orginally had the integral in the form of:
[itex]\frac{exp(x^{2})(x^{2})}{x^{3}}[/itex]-[itex]\frac{exp(x^{2})}{x^{3}}[/itex]

But I don't know how to integrate either!

Any help or suggexstions will be appreciated!
[itex]\displaystyle \frac{e^{(x^2)}(x^{2}-1)}{x^3}=\frac{e^{(x^2)}x^2}{x^3}-\frac{e^{(x^2)}}{x^3}[/itex]
[itex]\displaystyle =
\frac{e^{(x^2)}}{x}-\frac{e^{(x^2)}}{x^3}
[/itex]​
The substitution suggested by hogrampage is the place to start.

Then it looks like you will need to do integration by parts.

Added in Edit:

Well, yes, it can be done by inspection using the quotient rule in 'reverse'. Simply multiply the integrand by x/x.

[itex]\displaystyle =
\frac{1}{2}\frac{2xe^{(x^2)}x^{2}-e^{(x^2)}\,2x}{x^4}
[/itex]

I presume it can be done similarly using the product rule.
 
Last edited:
  • #4
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Thanks!
I know the answer is [itex]\frac{exp(x)}{2x}[/itex], becuase I can confirm it using the quotient rule for differentiation, but could someone please explain how to show the integration process?
 
  • #5
SammyS
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Thanks!
I know the answer is [itex]\frac{exp(x)}{2x}[/itex], becuase I can confirm it using the quotient rule for differentiation, but could someone please explain how to show the integration process?
Actually, that's not the result.

[itex]\displaystyle \frac{d}{dx}\left(\frac{\exp(x)}{2x}\right)=\frac{e^x (x-1)}{2 x^2}[/itex]

Earlier you said the result was [itex]\displaystyle \frac{\exp(x^2)}{2x}\ .[/itex]

That's not it either.

[itex]\displaystyle \frac{d}{dx}\left(\frac{\exp(x^2)}{2x}\right)= \frac{\exp(x^2)(2x^2-1)}{2x^2}[/itex]

Do you know integration by parts?
 
  • #6
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[tex]\int \frac{e^{x^2}(x^2-1)}{x^4} xdx[/tex] using the substitution suggested I got, [tex]\int \frac{e^{u}(u-1)}{2u^2} du[/tex]= [tex] \frac{e^{u}-1}{2u} [/tex].
Substituting back to x, [tex] \frac{e^{x^2}-1}{2x^2} [/tex]. The limits on the integration were from 0 to 2.
[tex] \frac{e^{4}-1}{8} - 0 = 6.6997... [/tex]
However, when I do this on wolfram, I get, [tex] \frac{e^{4}-5}{8} = 6.1997... [/tex]
I can't see what I am doing wrong.
 
  • #7
SammyS
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[tex]\int \frac{e^{x^2}(x^2-1)}{x^4} xdx[/tex] using the substitution suggested I got, [tex]\int \frac{e^{u}(u-1)}{2u^2} du[/tex]= [tex] \frac{e^{u}-1}{2u} [/tex]...

I can't see what I am doing wrong.
[tex]\int \frac{e^{u}(u-1)}{2u^2} du[/tex]≠ [tex] \frac{e^{u}-1}{2u} [/tex]

[itex]\displaystyle \frac{e^{u}(u-1)}{u^2}=\frac{u\cdot e^{u}-e^{u}\cdot 1}{u^2}[/itex]

Quotient rule:

[itex]\displaystyle \frac{\text{(Low) D(High)} - \text{(High) D(Low)}}{\text{(Low)}\!^2}[/itex]
 
  • #8
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Thanks Sammy,
Now I see that I made a mistake at the very beginning.
I wanted to integrate [tex]\int \frac{e^{x^2}(x^2-1)+1}{x^4} xdx[/tex] in which case I think the integral is correct, and still unsure where the 5 comes from?
 

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