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

[notnottrue]

Homework Statement

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

Homework Equations

Wolfram returns this answer
\frac{exp(x^{2})}{2x}

The Attempt at a Solution

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

But I don't know how to integrate either!

Any help or suggestions will be appreciated!

 

[hogrampage]

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.

 

[SammyS]

Homework Statement

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

Homework Equations

Wolfram returns this answer
\frac{exp(x^{2})}{2x}

The Attempt at a Solution

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

But I don't know how to integrate either!

Any help or suggexstions will be appreciated!

\displaystyle \frac{e^{(x^2)}(x^{2}-1)}{x^3}=\frac{e^{(x^2)}x^2}{x^3}-\frac{e^{(x^2)}}{x^3}

\displaystyle =<br /> \frac{e^{(x^2)}}{x}-\frac{e^{(x^2)}}{x^3}<br />​

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.

\displaystyle =<br /> \frac{1}{2}\frac{2xe^{(x^2)}x^{2}-e^{(x^2)}\,2x}{x^4}<br />

I presume it can be done similarly using the product rule.

 

[notnottrue]

Thanks!
I know the answer is \frac{exp(x)}{2x}, becuase I can confirm it using the quotient rule for differentiation, but could someone please explain how to show the integration process?

 

[SammyS]

Thanks!
I know the answer is \frac{exp(x)}{2x}, 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.

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

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

That's not it either.

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

Do you know integration by parts?

 

[notnottrue]

\int \frac{e^{x^2}(x^2-1)}{x^4} xdx using the substitution suggested I got, \int \frac{e^{u}(u-1)}{2u^2} du= \frac{e^{u}-1}{2u}.
Substituting back to x, \frac{e^{x^2}-1}{2x^2}. The limits on the integration were from 0 to 2.
\frac{e^{4}-1}{8} - 0 = 6.6997...
However, when I do this on wolfram, I get, \frac{e^{4}-5}{8} = 6.1997...
I can't see what I am doing wrong.

 

[SammyS]

\int \frac{e^{x^2}(x^2-1)}{x^4} xdx using the substitution suggested I got, \int \frac{e^{u}(u-1)}{2u^2} du= \frac{e^{u}-1}{2u}...

I can't see what I am doing wrong.

\int \frac{e^{u}(u-1)}{2u^2} du≠ \frac{e^{u}-1}{2u}

\displaystyle \frac{e^{u}(u-1)}{u^2}=\frac{u\cdot e^{u}-e^{u}\cdot 1}{u^2}

Quotient rule:

\displaystyle \frac{\text{(Low) D(High)} - \text{(High) D(Low)}}{\text{(Low)}\!^2}

 

[notnottrue]

Thanks Sammy,
Now I see that I made a mistake at the very beginning.
I wanted to integrate \int \frac{e^{x^2}(x^2-1)+1}{x^4} xdx in which case I think the integral is correct, and still unsure where the 5 comes from?