How to Convert an Integral from (1 + x) to 1 - n/(x+n)

  • Thread starter confused88
  • Start date
  • Tags
    Integrate
In summary, the person is having trouble understanding the integration of \int1/(1+x)2 dx with the range of the integral from 0 to x/n. They are suggested to use u substitution and compute dx/du. The final answer is achieved through an algebraic trick.
  • #1
confused88
22
0
Hi! I'm having trouble understanding my textbook, so can someone please explain to me how they got from


[tex]\int1/(1+x)[/tex]2 dx, with the range of the integral from 0 to x/n

to

1 - n/(x+n)


THank you So MuCh
 
Physics news on Phys.org
  • #2


Are you having trouble with the integration or the final answer?

u substitution works fine here, let u = (1 + x) and integrate. The final answer is a simple algebraic trick, nothing spectacular.
 
  • #3


I'm just having trouble with the integration ><. Oh wells
 
  • #4


You're looking for the integral of [tex]\mathop \smallint \nolimits_0^{x/n} \frac{1}{{(1 + x')^2 }}dx'[/tex] (technically you're not allowed to have your variable of integration in your bounds)

Make a substitution so that your integral now becomes [tex]\mathop \smallint \nolimits_0^{x/n} \frac{1}{{(u)^2 }}du\frac{{dx}}{{du}}[/tex] and remember to compute dx/du.

What, when you take its derivative becomes [tex]\frac{1}{{u^2 }}[/tex]? Find what that is, substitute back in for what you had set u to and you can plug in your integration limits and wala!
 

FAQ: How to Convert an Integral from (1 + x) to 1 - n/(x+n)

What does "U = (1 + x) and integrate" mean?

"U = (1 + x) and integrate" is a mathematical expression that represents a function, "U", which is equal to the sum of 1 and x, and the process of finding the area under the curve of this function.

How do you integrate U = (1 + x)?

To integrate U = (1 + x), you can use the power rule for integration, which states that the integral of x^n is equal to (x^(n+1))/(n+1). In this case, n is equal to 1, so the integral of x is (x^2)/2. Therefore, the integral of U = (1 + x) is (1/2)x^2 + C, where C is a constant.

What is the purpose of integrating U = (1 + x)?

The purpose of integrating U = (1 + x) is to find the area under the curve of this function. This is useful in many applications, such as in physics, engineering, and economics, where finding the total amount of something is important.

Can you provide an example of integrating U = (1 + x)?

For example, if we want to find the area under the curve of U = (1 + x) from x = 0 to x = 2, we can use the integral formula to calculate it. The integral of U = (1 + x) is (1/2)x^2 + C, so the area under the curve is ((1/2)2^2 + C) - ((1/2)0^2 + C) = 2 + C, where C is the constant of integration.

Is the integral of U = (1 + x) the same as the derivative of (1 + x)?

No, the integral of U = (1 + x) is not the same as the derivative of (1 + x). The integral is the reverse process of taking the derivative, so while the derivative of (1 + x) is equal to 1, the integral of U = (1 + x) is (1/2)x^2 + C. They are related, but not the same.

Similar threads

Back
Top