Proving Integral of (1+x^2)^n: Techniques and Examples

[Alexx1]

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

Can someone prove this?

 

[berkeman]

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

Can someone prove this?

What is the context of your question? Is it for schoolwork/homework?

 

[Alexx1]

What is the context of your question? Is it for schoolwork/homework?

I have exam (university) January 15th and we have exercises but we don't have answers.. so I would lik to know how to solve this one

 

[berkeman]

I have exam (university) January 15th and we have exercises but we don't have answers.. so I would lik to know how to solve this one

Okay. Schoolwork needs to go in the Homework Help forums, and you need to show some effort on trying to solve it. I'll move the thread now. Can you say anything about potential ways to solve the problem?

 

[l'Hôpital]

https://www.physicsforums.com/showthread.php?t=365687

Why did you make a new thread?

 

[yungman]

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

Can someone prove this?

Have you try x=tan(\theta)

1+tan^{2}(\theta)=sec^{2}(\theta)

 

[andylu224]

I think you made a mistake typing it: It should be '+' in between the 2 fraction and the integral.

Just do a simple Integration by parts without induction.

Let dv=dx , u = 1/(1+x²)ⁿ

I_n = x/(1+x²)ⁿ + 2n(integral)[x²/(1+x²)ⁿ⁺¹]dx

as 'x² = 1 + x² - 1',

You should end up with: I_n = x/(1+x²)ⁿ + 2n(I_n - I_n+1)

Make I_n+1 the subject. Finally lower each of the n terms by 1. (so n+1 -> n, and n -> n-1)