How do you prove this?

[Alexx1]

I_n = Integral(1/((1+x²)ⁿ)dx

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

Can someone prove this?

 

[HallsofIvy]

Use proof by induction on n, using integration by parts for the induction step.

 

[Alexx1]

Use proof by induction on n, using integration by parts for the induction step.

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)) ?

 

[Tedjn]

Let me rewrite your problem in TeX. Prove for all $n \in \mathbb{N}$ and $n \geq 2$ 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 I_n gives you an integral with a power of n+1 in the denominator, which you can't simplify. Instead, try expanding I_n-1 and working with that expression.

 

[Alexx1]

Let me rewrite your problem in TeX. Prove for all $n \in \mathbb{N}$ and $n \geq 2$ 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 I_n gives you an integral with a power of n+1 in the denominator, which you can't simplify. Instead, try expanding I_n-1 and working with that expression.

I don't know how to do that.. can you explain it to me?

 

[Tedjn]

What does integration by parts on I_n-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.

 

[Alexx1]

What does integration by parts on I_n-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.

I absolutely have no idea how to integrate it by parts on I_n-1..

 

[Jame]

I absolutely have no idea how to integrate it by parts on I_n-1..

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.

 

[Alexx1]

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.

Is it like: x^n-1 = xⁿ*x^-1 ?

 

[Jame]

Is it like: x^n-1 = xⁿ*x^-1 ?

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 $$\int \frac{dx}{(1+x^2)^{n-1}}$$, 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.

 

[Alexx1]

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 $$\int \frac{dx}{(1+x^2)^{n-1}}$$, 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.

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 I_n and the result must be $$\int \frac{dx}{(1+x^2)^n} \;=\; \frac{1}{2(n-1)}\cdot\frac{x}{(1+x^2)^{n-1}} \;\;-\;\; \frac{2n-3}{2(n-1)}\int \frac{dx}{(1+x^2)^{n-1}}$$