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x^3\(x^8-2)的不定积分为多少
题目详情
x^3\(x^8-2)的不定积分为多少
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答案和解析
1/(x^8-2)=1/[(x^4-√2）*（x^4+√2)]
=√2/4*[1/(x^4-√2)-1/(x^4+√2)]
∫x^3/(x^8-2) dx
=∫1/(x^8-2) d(x^4/4)
=1/4*∫1/(x^8-2) d(x^4)
=1/4*∫√2/4*[1/(x^4-√2)-1/(x^4+√2)] d(x^4)
=√2/16*∫[1/(x^4-√2)-1/(x^4+√2)] d(x^4)
=√2/16*ln|x^4-√2|-√2/16*ln|x^4+√2| +C
=√2/4*[1/(x^4-√2)-1/(x^4+√2)]
∫x^3/(x^8-2) dx
=∫1/(x^8-2) d(x^4/4)
=1/4*∫1/(x^8-2) d(x^4)
=1/4*∫√2/4*[1/(x^4-√2)-1/(x^4+√2)] d(x^4)
=√2/16*∫[1/(x^4-√2)-1/(x^4+√2)] d(x^4)
=√2/16*ln|x^4-√2|-√2/16*ln|x^4+√2| +C
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