Reduction Formulae: Defining I(n) & Establishing Reduction Formula

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SUMMARY

The discussion focuses on defining the integral I(n) as I(n) = ∫(x^n / √(1+x^2)) dx. The evaluations of I(0) and I(1) yield ln(1+√2) and ln(√2), respectively. A reduction formula is established as I(n) = (√2 - (n-1)I(n-2)) / n for suitable values of n. Participants suggest using integration by parts to derive the reduction formula, specifically by letting dv = (x/√(1+x^2)) dx and u = x^(n-1).

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  • Understanding of integral calculus, specifically integration techniques.
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  • Knowledge of logarithmic functions and their properties.
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Badrakhandama
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Define I(n) by I(n) = integral of x^n / square root(1+x^2) dx. Evaluate I(0), I(1) and then establish the reduction formula I(n) = ((root2)-(n-1)I(n-2))/(n) for suitable values of n, which should be stated




Here is my attempt


I found I(0) first,and got the answer to be ln(1+root2), and I(1) to be ln (root2).

For the reduction formula, I am not sure where to start - do I try it by parts, differentiating x^n, and integrating 1/(root(1+x^2)) , or what should I do? For the values of n, I am not sure either.
 
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Try the following for integration by parts & see where it leads.

\text{Let }\ dv=\frac{x}{\sqrt{1+x^2}}\,dx\ \text{ and let }\ u=x^{n-1}.

That should result in an integral with \sqrt{1+x^2} in the numerator.

Multiply the integrand by \frac{\sqrt{1+x^2}}{\sqrt{1+x^2}}\ , and simplify the numerator.
 

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