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Q. `int` `(x^6 - 1)/(x^2 + 1)` `dx`   = ?

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user8235304 | Student, Grade 11 | Valedictorian

Posted July 13, 2013 at 3:25 AM via web

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Q. `int` `(x^6 - 1)/(x^2 + 1)` `dx`   = ?

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jeew-m | College Teacher | (Level 1) Educator Emeritus

Posted July 13, 2013 at 3:51 AM (Answer #1)

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Using long division it can be obtained that;

`(x^6-1)/(x^2+1) = (x^4-x^2+1)-2/(x^2+1)`

 

`int(x^6-1)/(x^2+1)dx `

`= int((x^4-x^2+1)-2/(x^2+1))dx`

`= int(x^4-x^2+1)dx-2int(1/(x^2+1))dx`

`= x^4/4-x^3/3+x-2tan^(-1)x+C` where C is constant

 

So the answer is

`int(x^6-1)/(x^2+1)dx` = `x^4/4-x^3/3+x-2tan^(-1)x+C`

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