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How do you prove this?

  1. Dec 27, 2009 #1
    In = Integral(1/((1+x2)n)dx

    For all n ∈ Natural numbers (exclusively 0), n≠1: In = 1/(2(n-1)) * x/((1+x2)n-1) + (2n-3)/(2(n-1))*In-1

    Can someone prove this?
    Last edited: Dec 27, 2009
  2. jcsd
  3. Dec 27, 2009 #2


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    Use proof by induction on n, using integration by parts for the induction step.
  4. Dec 27, 2009 #3
    I've tried it with that methode..
    But what do you choose as u and dv?

    I've tried it with: u=(1/((1+x2)n) ==> du= (-2nx)/((1+x2)n+1) and dv= 1 ==> v=x

    ==> x/(1+x2)n + 2n Integral(x^2/((1+x^2)^(n+1)))

    How do you integrate: (x^2/((1+x^2)^(n+1)) ?
  5. Dec 27, 2009 #4
    Let me rewrite your problem in TeX. Prove for all [itex]n \in \mathbb{N}[/itex] and [itex]n \geq 2[/itex] that

    \int \frac{dx}{(1+x^2)^n} \;=\; \frac{1}{2(n-1)}\cdot\frac{x}{(1+x^2)^{n-1}}
    \;\;-\;\; \frac{2n-3}{2(n-1)}\int \frac{dx}{(1+x^2)^{n-1}}

    You've noticed that using integration by parts on In gives you an integral with a power of n+1 in the denominator, which you can't simplify. Instead, try expanding In-1 and working with that expression.
  6. Dec 28, 2009 #5
    I don't know how to do that.. can you explain it to me?
  7. Dec 28, 2009 #6
    What does integration by parts on In-1 give you? The idea is that you might be able to manipulate it in some way to achieve the desired form, because all the terms turn out to be of the right form.
  8. Dec 28, 2009 #7
    I absolutely have no idea how to integrate it by parts on In-1..
  9. Dec 28, 2009 #8
    Think what the expression means! You have n-1 instead of n, so what? n is any number you want anyway. Call it m if that makes it easier.
  10. Dec 29, 2009 #9
    Is it like: xn-1 = xn*x-1 ?
  11. Dec 29, 2009 #10
    I know for sure that you are not as stupid as you're making yourself look right now, it's actually a classic mistake: to think the math is over your head and give up the logic and instead blindly follow what people tell you. Enough rant, sorry.

    You have already been given the solution. Look
    \int \frac{dx}{(1+x^2)^n} \;=\; \frac{1}{2(n-1)}\cdot\frac{x}{(1+x^2)^{n-1}}
    \;\;-\;\; \frac{2n-3}{2(n-1)}\int \frac{dx}{(1+x^2)^{n-1}}

    Put a big and fat parenthesis around [tex]\int \frac{dx}{(1+x^2)^{n-1}}[/tex], or even write it separately. This is the exact same problem! Solve it again and you get an integral with n-2, again and you get n-3, and so on. Try setting n = 3 and do this two times to see the point.
  12. Dec 29, 2009 #11
    I knew the answer in the beginning! The only thing I have to do is prove it!
    Not proving it by filling in a random figure!
    I have to solve integral In and the result must be [tex]
    \int \frac{dx}{(1+x^2)^n} \;=\; \frac{1}{2(n-1)}\cdot\frac{x}{(1+x^2)^{n-1}}
    \;\;-\;\; \frac{2n-3}{2(n-1)}\int \frac{dx}{(1+x^2)^{n-1}}
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