Question regarding an integral

[Christoffelsymbol100]

Homework Statement

I've got an equation which I need to integrate. However, integrating it and checking with the solutions, I get two different results. I get the same result as using wolfram alpha, but a different result from the book.
If I differentiate both results, I get back to the orginal equation, so I don't know why they are different.

Homework Equations

Equation which I need to Integrate by separation of variables: \frac{d\dot{x}^2}{dx} = \frac{2l^4 a^2 x}{(12x^2+l^2)^2}

Result from Book: \dot{x}^2 = \frac{ l^2 a^2 x^2}{12x^2+l^2}
Result from Wolfram-Alpha: \dot{x}^2 = -\frac{(a^2 l^4)}{(12 (l^2+12 x^2))}

The Attempt at a Solution

If I differentiate both, I'll both get the same equation back which I integrated, I just don't understand why that's the case, since it doesn't look like they differ by a additive constant.

 

[Ray Vickson]

Homework Statement

I've got an equation which I need to integrate. However, integrating it and checking with the solutions, I get two different results. I get the same result as using wolfram alpha, but a different result from the book.
If I differentiate both results, I get back to the orginal equation, so I don't know why they are different.

Homework Equations

Equation which I need to Integrate by separation of variables: \frac{d\dot{x}^2}{dx} = \frac{2l^4 a x}{(12x^2+l^2)^2}

Result from Book: \dot{x}^2 = \frac{ l^2 a^2 x^2}{12x^2+l^2}
Result from Wolfram-Alpha: \dot{x}^2 = -\frac{(a^2 l^4)}{(12 (l^2+12 x^2))}

The Attempt at a Solution

If I differentiate both, I'll both get the same equation back which I integrated, I just don't understand why that's the case, since it doesn't look like they differ by a additive constant.

(1) Please clarify: does
\frac{d\dot{x}^2}{dx}
mean
\frac{d( d(x^2)/dt)}{dx}
or
\frac{d (dx/dt)^2}{dx}
or
\left( \frac{d (dx / dt) }{dx}\right)^2 \:?
(2) Show the steps you used in obtaining your answer.

 

[Christoffelsymbol100]

It means :\frac{d\dot{x}^2}{dx} = \frac{d}{dx}\left(\frac{dx}{dt}\right)^2

For the result from the book there where no steps shown.

For my result: \frac{d\dot{x}^2}{dx} = \frac{2l^4a^2x}{(12x^2+l^2)^2} \Longrightarrow \int d\dot{x}^2 =2 l^4a^2\int \frac{x}{(12x^2+l^2)^2} dx

Apply Substitution: u = 12x^2 +l^2 \,\,\,\,\,\,\, \frac{du}{dx} = 24x \Rightarrow dx = \frac{1}{24x}du

\int d\dot{x}^2 = \dot{x}^2 = l^4a^2 \int\frac{1}{24u^2}du = -2l^4a^2\frac{1}{24} u^{-1} + c = -l^4a^2\frac{1}{12}\frac{1}{12x^2+l^2}+c

There were some starting conditions which would lead to c = 0

 

[Ray Vickson]

It means :\frac{d\dot{x}^2}{dx} = \frac{d}{dx}\left(\frac{dx}{dt}\right)^2

For the result from the book there where no steps shown.

For my result: \frac{d\dot{x}^2}{dx} = \frac{2l^4a^2x}{(12x^2+l^2)^2} \Longrightarrow \int d\dot{x}^2 =2 l^4a^2\int \frac{x}{(12x^2+l^2)^2} dx

Apply Substitution: u = 12x^2 +l^2 \,\,\,\,\,\,\, \frac{du}{dx} = 24x \Rightarrow dx = \frac{1}{24x}du

\int d\dot{x}^2 = \dot{x}^2 = l^4a^2 \int\frac{1}{24u^2}du = -2l^4a^2\frac{1}{24} u^{-1} + c = -l^4a^2\frac{1}{12}\frac{1}{12x^2+l^2}+c

There were some starting conditions which would lead to c = 0

You are correct, and the book is wrong (as can be checked by differentiating the answer---a step you should always perform).

 

[Christoffelsymbol100]

I did perform that step as I wrote in my post and I get the same equation, though my derivative might be wrong.

Let's differentiate the solution from the book with respect to x:

\frac{d}{dx}\left(\frac{l^2a^2x^2}{12x^2+l^2}\right) = l^2a^2 \frac{d}{dx}\left(\frac{x^2}{12x^2+l^2}\right)

Using the quotient rule I get: l^2a^2 \frac{2x(12x^2+l^2)-24x^3}{(12x^2+l^2)^2} = l^2a^2\frac{24x^3-2xl^2-24x^3}{(12x^2+l^2)^2} = \frac{2l^4a^2x}{(12x^2+l^2)^2} which is the same as the orginal function which I integrated

 

[vela]

Even though it doesn't look like it at first glance, the two results do, in fact, differ by just an additive constant:
$$\frac{a^2l^2}{12}-\frac{a^2l^4}{12(12x^2+l^2)} = \frac{l^2a^2x^2}{12x^2+l^2}$$

 

[Ray Vickson]

Even though it doesn't look like it at first glance, the two results do, in fact, differ by just an additive constant:
$$\frac{a^2l^2}{12}-\frac{a^2l^4}{12(12x^2+l^2)} = \frac{l^2a^2x^2}{12x^2+l^2}$$

Oops, my mistake in #4.

 

[Christoffelsymbol100]

Ah I see, thanks very much both of you!