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Im having trouble understanding questions involving deriving a recursion formula.

I need to derive the recursion formula for $I_n$ where $n>=2$ $$I_n = \int(x^2-1)^n dx$$ The other questions ive done so far I used the integration by parts and derived a formula where n is decreasing in the integral. I am using the same method and I got this answer: $$x(x^2-1)^n-2n\int x^2(x^2-1)^{n-1}$$ However the textbook is giving me a different answer.

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up vote 3 down vote accepted

Use integration by parts, letting $u=(x^2-1)^n$, and $dv=dx$.

Then $du=2nx(x^2-1)^{n-1}\,dx$ and we can take $v=x$. Thus $$I_n=x(x^2-1)^n -2n\int x^2(x^2-1)^{n-1}\,dx.$$ Note that $$x^2(x^2-1)^{n-1}=\left((x^2-1)+1\right)(x^2-1)^{n-1}=(x^2-1)^{n}+(x^2-1)^{n-1}.$$ We get $$I_n=x(x^2-1)^{n} -2nI_n -2nI_{n-1},$$ and can solve for $I_n$ in terms of $I_{n-1}$.

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Im confused when you are rewriting the integrand. –  Ddayne Feb 26 '13 at 5:43
    
I have rewritten the derivation of that part, hope it is clear now. –  André Nicolas Feb 26 '13 at 5:50
    
I feel dumb for asking but why does (x^2-1+1)(x^2-1)^n-1 become (x^2-1)^n + (x^2-1)^n-1? –  Ddayne Feb 26 '13 at 6:39
    
$[(x^2-1)+1]A=(x^2-1)A+A$. Here $A=(x^2-1)^{n-1}$. –  André Nicolas Feb 26 '13 at 6:41
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