Integrating dz/(z^2+2z/x): Step-by-Step Guide and Troubleshooting Tips

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To integrate dz/(z^2 + 2z/x), partial fractions can be used, but completing the square is recommended for a more straightforward approach. The integral can be expressed in terms of hyperbolic trigonometric functions, which also leads to logarithmic solutions. The problem features simple poles at z=0 and z=-2/x, affecting the integration process and domain restrictions. Both methods ultimately yield the same solution within the common domain of validity, specifically for z in the range (-2/x, 0) when x is positive. Proper attention to the integration technique ensures accurate results.
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Homework Statement


How to integrate dz/(z^2+2z/x)?

Homework Equations


None.

The Attempt at a Solution


I did partial fractions but it doesn't seem to work.
A/z+B/(z+2/x)=1
 
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Try completing the square!
 
The thing to do for partial fractions would be to solve:
1/(z^2+2z/x) = 1 / z(z+2/x) = A/z+B/(z+2/x)
for A and B, and then it can be integrated to logs.

Or, you can complete the square and you're left with an integral that yields a hyperbolic trig solution which can be expressed in terms of logs:
http://en.wikipedia.org/wiki/Hyperbolic_function

The interesting thing about this problem is that it has simple poles at z=0 and -2/x.
So you get log absolute values when integrating 1/z at negative z...,
and there are corresponding domain restrictions on the validity of the hyperbolic trig solution.
When you take care, you can arrive at the two methods yielding the same solution over their common domain of validity: z=(-2/x,0) (assuming x>0).
 
Math10 said:
I did partial fractions but it doesn't seem to work.
A/z+B/(z+2/x)=1
That would be because the righthand side shouldn't be equal to 1.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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