What Are the Restricted x Values for the Integral of dx/(sqrt(d^2 + x^2))?

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In summary, the conversation discusses the evaluation of an expression and the resulting difference between two solutions. The speaker also asks about the meaning of restricted values and where a constant in the solution came from. The summary concludes that the result is valid for any values of x and y as long as x^2+y^2 does not equal 0, and that the constant 'd' represents a distance in the problem.
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carlosbgois
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Hi there. Evaluating the expression [itex]\int\frac{dx}{\sqrt{x^{2}+y^{2}}}[/itex] I can get to the result [itex]ln(\frac{x+\sqrt{x^{2}+y^{2}}}{d})[/itex], but in my book it goes from this directly to [itex]ln (x+\sqrt{x^{2}+y^{2}})[/itex], a result wolframalpha says is valid for 'restricted [itex]x[/itex] values'. What does it mean? What are those restricted values? Why?

Many thanks.
 
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  • #2
carlosbgois said:
Hi there. Evaluating the expression [itex]\int\frac{dx}{\sqrt{x^{2}+y^{2}}}[/itex] I can get to the result [itex]ln(\frac{x+\sqrt{x^{2}+y^{2}}}{d})[/itex], but in my book it goes from this directly to [itex]ln (x+\sqrt{x^{2}+y^{2}})[/itex], a result wolframalpha says is valid for 'restricted [itex]x[/itex] values'. What does it mean? What are those restricted values? Why?

Many thanks.



Well, since this is indefinite integration both the results are correct as their difference is just the constant [itex]-\ln d[/itex].

The question is: where did you get the constant [itex]d[/itex] from??

The result is valid for any values of [itex]x,y, s.t. x^2+y^2\neq 0[/itex]

DonAntonio
 
  • #3
Thank you. 'd' is actually a constant length (the distance from a point p to a disk, in an axis that goes through the center of the disk)
 

FAQ: What Are the Restricted x Values for the Integral of dx/(sqrt(d^2 + x^2))?

What is the meaning of the integral of dx/(sqrt(d^2+x^2))?

The integral of dx/(sqrt(d^2+x^2)) represents the area under the curve of the function 1/(sqrt(d^2+x^2)) with respect to the variable x. It is a mathematical concept used to calculate the total value of a function within a given range.

2. How do you solve the integral of dx/(sqrt(d^2+x^2))?

The integral of dx/(sqrt(d^2+x^2)) can be solved using the substitution method, where u = d^2+x^2 and dx = (du-2x)/2. After substitution, the integral becomes 1/(sqrt(u))du, which can be easily integrated using the power rule. Once the integral is solved, the substitution must be undone to get the final solution in terms of x.

3. What is the domain of the integral of dx/(sqrt(d^2+x^2))?

The domain of the integral of dx/(sqrt(d^2+x^2)) is the set of all real numbers except for x = -d, as the function is undefined at this point due to division by zero. This means that the integral can be evaluated for all values of x, except for -d.

4. Why is the integral of dx/(sqrt(d^2+x^2)) important in mathematics?

The integral of dx/(sqrt(d^2+x^2)) has many applications in various fields of mathematics, including calculus, physics, and engineering. It is used to calculate the area under a curve, which is essential in finding the total value of a function and solving problems related to motion, energy, and other physical phenomena.

5. Can the integral of dx/(sqrt(d^2+x^2)) be evaluated using numerical methods?

Yes, the integral of dx/(sqrt(d^2+x^2)) can be evaluated numerically using methods such as the trapezoidal rule or Simpson's rule. These methods approximate the integral by dividing the given range into smaller intervals and calculating the area under the curve for each interval. These approximations become more accurate as the number of intervals increases.

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