Integrate (x2 - 1)/(x2 + 1): Lunar Guy's Question

In summary, the conversation discusses integrating the function (x2 - 1)/(x2 + 1) on the interval [0,1], with the final result being -x - arctan(x). The process involves canceling out terms, converting the numerator to the form of the denominator, and plugging in the numbers to check for correctness.
  • #1
Lunar Guy
17
0
1. Integrate on the interval [0, 1]: (x2 - 1)/(x2 + 1)

2. No relevant equations.

3.

S x2/(x2 + 1) - [S 1/(x2 + 1)] =

S (x2 + 1) - 1/(x2 + 1) - [(arctan x)] =

S -1 - [(arctan x)] =

-x - arctan x

I can plug in the numbers on my own, but I would like to know if what I've done so far is correct...

Thanks to anyone who reads this and answers it.

- Lunar Guy
 
Last edited:
Physics news on Phys.org
  • #2
How does ((x2 + 1) - 1)/(x2 + 1) become -1?
 
  • #3
the (x^2 + 1) cancles
but where did u get x2/(x2 + 1) from?
x^2 - 1 = x - 1)(x+1)
 
  • #4
a.a said:
the (x^2 + 1) cancles
but where did u get x2/(x2 + 1) from?
x^2 - 1 = x - 1)(x+1)

The (x2 + 1) cancels to become 1 - (1/(x2 + 1)), not -1.
He got x2/(x2 + 1) from (a + b)/c = a/c + b/c.
 
  • #5
so how does x^2/(x^2 +1) reduce to x^2 +1?
 
  • #6
a.a said:
so how does x^2/(x^2 +1) reduce to x^2 +1?
That never happened. He skipped the step x2 = x2 + 1 - 1.
 
  • #7
Think its quite easy... Convert the neumerator in the form of the denominator by writing -1 as 1-2
 
  • #8
So now you get (xsquare + 1) - 2
 

1. What is the process for integrating (x2 - 1)/(x2 + 1)?

The process for integrating (x2 - 1)/(x2 + 1) involves using the substitution method. First, substitute x with tan(u) and dx with sec2(u)du. Then, use algebraic manipulation and the trigonometric identity sec2(u) = 1 + tan2(u) to simplify the expression. Finally, integrate the simplified expression and substitute back in the original variable of x.

2. Can (x2 - 1)/(x2 + 1) be integrated using any other method?

Yes, (x2 - 1)/(x2 + 1) can also be integrated using the partial fractions method. This involves breaking the fraction into smaller, simpler fractions and integrating each one separately.

3. Is the integration of (x2 - 1)/(x2 + 1) a difficult process?

The difficulty of integrating (x2 - 1)/(x2 + 1) depends on the method used and the individual's familiarity with that method. Some may find the substitution method easier, while others may prefer the partial fractions method.

4. Can you give an example of a real-life application of integrating (x2 - 1)/(x2 + 1)?

Integrating (x2 - 1)/(x2 + 1) has many real-life applications, such as in physics, engineering, and economics. For example, it can be used to calculate the area under a curve in a velocity-time graph to determine the distance traveled by an object.

5. Is there a way to check if the integration of (x2 - 1)/(x2 + 1) is correct?

Yes, the integration of (x2 - 1)/(x2 + 1) can be checked by taking the derivative of the integrated expression. If the derivative is equal to the original function, then the integration is correct.

Similar threads

  • Calculus and Beyond Homework Help
Replies
3
Views
273
  • Calculus and Beyond Homework Help
Replies
11
Views
1K
  • Calculus and Beyond Homework Help
Replies
8
Views
1K
  • Calculus and Beyond Homework Help
Replies
14
Views
2K
  • Calculus and Beyond Homework Help
Replies
3
Views
1K
  • Calculus and Beyond Homework Help
Replies
10
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
345
  • Calculus and Beyond Homework Help
Replies
6
Views
548
  • Calculus and Beyond Homework Help
Replies
15
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
1K
Back
Top