∫1/(x^4-1)dx

发布网友发布时间：2022-11-28 07:53

共3个回答

热心网友时间：2023-10-22 08:00

1/(x^4-1)
=1/(x²+1)(x²-1)
=1/2*[1/x²-1)-1/x²+1)]

同理1/(x²-1)=1/2*[1/(x-1)-1/(x+1)]
所以1/(x^4-1)
=(1/4)*1/(x-1)-(1/4)*1/(x+1)-(1/2)*1/(x²+1)

所以原式=1/4∫dx/(x-1)-1/4∫dx/(x+1)-1/2∫dx/(x²+1)
=1/4∫d(x-1)/(x-1)-1/4∫d(x+1)/(x+1)-1/2∫dx/(x²+1)
=1/4*ln|x-1|-1/4*ln|x+1|-1/2*arctanx+C
=1/4*ln|(x-1)/(x+1)|-1/2*arctanx+C

热心网友时间：2023-10-22 08:01

-0.5*(arctanh(x)+arctan(x))+C

热心网友时间：2023-10-22 08:01

解：∫1/(x^4-1)dx
=∫1/{1/2[1/(x^2-1)-1/(x^2+1)]}dx
=1/2∫1/[1/(x^2-1)-1/(x^2+1)]dx
=1/2∫1/{1/2[/(x-1)-1/(x+1)]-1/(x^2+1)}dx
=1/2*1/2*ln(x-1)-1/2*1/2*ln(x+1)-2*arctanx+C1
=1/4ln(x-1)-1/4ln(x+1)-2arctanx+C1
=1/4ln[(x-1)/(x+1)]-2arctanx+C

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