ii. The sum of two cubes factorization can be used to see x^3 + 1 = (x + 1)(x^2 ? x + 1). Therefore, ? 1/2 0 1/(x^2 ? x + 1) dx = ? 1/2 0 (x + 1)/(x + 1)(x 2 ? x + 1) dx = ? 1/2 0 (x + 1)/( 1 + x^3) dx = ? 1/2 0 (x + 1)/( 1/(1 + x^3) ) dx. Use what weve talked about in class to find a series for 1/(1 + x^3) and use it above to obtain a series th

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ii. The sum of two cubes factorization can be used to see x^3 + 1 = (x + 1)(x^2 ? x + 1). Therefore, ? 1/2 0 1/(x^2 ? x + 1) dx = ? 1/2 0 (x + 1)/(x + 1)(x 2 ? x + 1) dx = ? 1/2 0 (x + 1)/( 1 + x^3) dx = ? 1/2 0 (x + 1)/( 1/(1 + x^3) ) dx. Use what weve talked about in class to find a series for 1/(1 + x^3) and use it above to obtain a series that equals the definite integral.

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Expert Answer to - ii. The sum of two cubes factorization can be used to see x^3 + 1 = (x + 1)(x^2  x + 1). Therefor

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Solution for - ii. The sum of two cubes factorization can be used to see x^3 + 1 = (x + 1)(x^2  x + 1). Therefor

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This an additional answer to - ii. The sum of two cubes factorization can be used to see x^3 + 1 = (x + 1)(x^2  x + 1). Therefor

(Solved): ii. The sum of two cubes factorization can be used to see x^3 + 1 = (x + 1)(x^2 x + 1). Therefor ...

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