# An integarl problem

1. Jul 1, 2010

### TrifidBlue

How can this be proved?

I really in need to know it, as I'm working on a derivation of Kepler's laws, but I'm stuck at this point ....

2. Jul 1, 2010

### Tedjn

To solve the integral, I would have tried to complete the square in the denominator before using trig substitution. However, due to the fundamental theorem of calculus, you might find it easier to take the derivative of the right hand side. There does seem to be an arbitrary constant missing.

3. Jul 1, 2010

### Gerenuk

This can be solved with the Abel substitution
$$t=(\sqrt{a+bx+cx^2})'$$

4. Jul 1, 2010

### Gerenuk

I wonder what's the easiest way :) Maybe you calculate and tell me ;)
Hint: For the Abel substitution calculate 4t^2Y and rearrange for Y (Y=a+bx+cx^2). Then solve with usual trig substitution. This also works if the square root is any other half-integer power.

In general for simple square roots you can also use

Last edited by a moderator: May 4, 2017
5. Jul 2, 2010

### Gib Z

Case 1: $b^2 - 4ac = 0$ in which case the quadratic term is a perfect square of a linear term, so the integral is simply the log of that term.

Case 2: If $b^2- 4ac > 0$, then completing the square and a simple linear shift will make the integrand of the form $\frac{1}{\sqrt{x^2 - a^2}}$, and x= a cosh t makes that one come out immediately.

Case 3: If $b^2-4ac < 0$, completing the square and a linear shift makes the integrand of the form $\frac{1}{\sqrt{x^2 + a^2}}$, and x= a sinh t makes that one come out immediately.

Of course the appropriate circular trigonometric substitutions would work as well, but don't come out as quickly. For both of these integrals, the substitutions lead straight to $\int dt$

6. Jul 3, 2010

### TrifidBlue

thank you very much Tedjn, Gerenuk and Gib Z
and extra thanks for you Gib Z for the details