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For n consider integral sign (upper limit3-lower limit 2) x^n/(x^2-1)dx. Show the...

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snsett | Student, Undergraduate | eNoter

Posted August 20, 2012 at 5:14 PM via web

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For n consider integral sign (upper limit3-lower limit 2) x^n/(x^2-1)dx.

Show the In+2-In=3^(n+1)-2^(n+1)/n+1

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted August 20, 2012 at 5:27 PM (Answer #1)

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You need to consider that `I_n = int_2^3 (x^n)/(x^2-1) dx` , hence, you may write `I_(n+2)`  such that:

`I_(n+2) = int_2^3 (x^(n+2))/(x^2-1) dx`

You need to evaluate `I_(n+2) - I_n`  such that:

`I_(n+2) - I_n = int_2^3 (x^(n+2))/(x^2-1) dx - int_2^3 (x^n)/(x^2-1) dx`

`I_(n+2) - I_n = int_2^3 (x^(n+2) - x^n)/(x^2-1) dx`

Notice that `x^(n+2) = x^n*x^2` , hence, you may factor out x^n such that:

`I_(n+2) - I_n = int_2^3 x^n*(x^2 - 1)/(x^2-1) dx`

Reducing by `x^2 - 1`  yields:

`I_(n+2) - I_n = int_2^3 x^n dx`

`I_(n+2) - I_n = (x^(n+1))/(n+1)|_2^3`

You need to remember that `int_a^b f(x)= F(b) - F(a).`

`I_(n+2) - I_n = 3^(n+1)/(n+1)- 2^(n+1)/(n+1) `

`I_(n+2) - I_n = (3^(n+1)- 2^(n+1))/(n+1) `

Hence, evaluating the given difference yields that `I_(n+2) - I_n = (3^(n+1)- 2^(n+1))/(n+1) .`

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