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

Homework Statement

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

Homework Equations

$\frac{exp(x^{2})}{2x}$

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 dont know how to integrate either!

Any help or suggestions will be appreciated!

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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.

SammyS
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Homework Statement

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

Homework Equations

$\frac{exp(x^{2})}{2x}$

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!
$\displaystyle \frac{e^{(x^2)}(x^{2}-1)}{x^3}=\frac{e^{(x^2)}x^2}{x^3}-\frac{e^{(x^2)}}{x^3}$
$\displaystyle = \frac{e^{(x^2)}}{x}-\frac{e^{(x^2)}}{x^3}$​
The substitution suggested by hogrampage is the place to start.

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

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

$\displaystyle = \frac{1}{2}\frac{2xe^{(x^2)}x^{2}-e^{(x^2)}\,2x}{x^4}$

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

Last edited:
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
Staff Emeritus
Homework Helper
Gold Member
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?

$$\int \frac{e^{x^2}(x^2-1)}{x^4} xdx$$ 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
Staff Emeritus
Homework Helper
Gold Member
$$\int \frac{e^{x^2}(x^2-1)}{x^4} xdx$$ 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:

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

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?
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