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If `I_n = int(1/(a+x^2)^n)dx` show that; `I_(n+1) = x/(2an(a+x^2)^n)+(2n-1)/(2an)I_n`...

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lak-86 | Student, Undergraduate | Salutatorian

Posted July 8, 2013 at 4:58 AM via web

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If `I_n = int(1/(a+x^2)^n)dx` show that;

`I_(n+1) = x/(2an(a+x^2)^n)+(2n-1)/(2an)I_n`

`n>=1`

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aruv | High School Teacher | Valedictorian

Posted July 8, 2013 at 9:25 AM (Answer #1)

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`I_n=int1/(a+x^2)^ndx`

`` Integrating by parts ,considering  first function as `1/(a+x^2)^n` and 1 as second function

`I_n={1/(a+x^2)^n}x-int{-n(a+x^2)^{-n-1}(2x)(x)dx`

`=x/(a+x^2)^n+2n int x^2/(a+x^2)^{n+1}dx`

`` `=x/(a+x^2)^n+2n int(a+x^2-a)/(a+x^2)^{n+1}dx`

`=x/(a+x^2)^n+2nint(a+x^2)/(a+x^2)^{n+1}dx-2ninta/(a+x^2)^{n+1}dx`

`=x/(a+x^2)^n+2nint1/(a+x^2)^ndx-2naint1/(a+x^2)^{n+1}dx`

`I_n=x/(a+x^2)^n+2nI_n-2naI_{n+1}`

`2anI_{n+1}=-I_n+2nI_n+x/(a+x^2)^n`

`I_{n+1}={(2n-1)/(2an)}I_n+x/{(2an)(a+x^2)^n}```

`I_{n+1}=x/{(2an)(a+x^2)^n}+(2n-1)/(2an)I_n`

where

`I_{n+1}=int1/(a+x^2)^{n+1}dx`

Thus we have proved the result.

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