How  to integerate the below equation: 1/(1 + x^6)dx.

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You should use the following formula such that:

`1 + x^6 = 1 + (x^2)^3 = (1 + x^2)(1 - x^2 + x^4)`

You need to use partial fraction decomposition such that:

`1/(1 + x^6) = (Ax+B)/(1+x^2) + (Cx+D)/(1 - x^2 + x^4)`

Bringing the fractions to a common denominator yields:

`1 = (Ax+B)(1 - x^2 + x^4) + (Cx+D)(1+x^2)`

`1 = Ax+B-Ax^3-Bx^2+Ax^5+Bx^4+Cx+D+Cx^3+Dx^2`

`1 = Ax^5 + Bx^4 + x^3(C-A) + x^2(D-B) + x(A+C) + B + D`

Equating the coefficients of like powers yields:

`A=0` 

`C-A=0 => C=A=0`

`D - B = 0`

`B + D = 1 => 2D = 1 => D=1/2 => B=1/2`

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

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

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

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

Hence, evaluating the given integral yields `int 1/(1 + x^6) dx = (1/2)tan^(-1)x + (1/2)int 1/(1 - x^2 + x^4)dx + c`

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