Integrate x/sqrt(a^2+b^2-2abx)

  • Thread starter Thread starter albega
  • Start date Start date
  • Tags Tags
    Integrate
albega
Messages
74
Reaction score
0

Homework Statement


I need to integrate x/sqrt(a^2+b^2-2abx) with respect to x.

The Attempt at a Solution


This follows from a substitution of the form x=cost in a textbook I'm reading - they jump to the solution straight from the above and I have no idea how to go about it - any hints will be very gratefully appreciated, thanks :)
 
Physics news on Phys.org
albega said:

Homework Statement


I need to integrate x/sqrt(a^2+b^2-2abx) with respect to x.

The Attempt at a Solution


This follows from a substitution of the form x=cost in a textbook I'm reading - they jump to the solution straight from the above and I have no idea how to go about it - any hints will be very gratefully appreciated, thanks :)

Hint: \int \frac{Au + B}{\sqrt u}\,du = \int Au^{1/2} + Bu^{-1/2}\,du.
 
pasmith said:
Hint: \int \frac{Au + B}{\sqrt u}\,du = \int Au^{1/2} + Bu^{-1/2}\,du.

Got it, thanks! Was that something you could spot or did you just know it from experience - I don't think I would ever have thought of something like that...
 
You also should make an attempt to post HW in the correct HW forum. There is a perfectly good Calculus-HW forum listed right below this one, which is where this post belongs.
 
SteamKing said:
You also should make an attempt to post HW in the correct HW forum. There is a perfectly good Calculus-HW forum listed right below this one, which is where this post belongs.

Thread moved to Calculus HH. :smile:
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top