X的四次幂加一分之一的不定积分怎么运算
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∫1/(x^4+1)dx
=1/2∫[(x²+1)-(x²-1)]/(x^4+1)dx
=1/2[∫(x²+1)/(x^4+1)dx-∫(x²-1)/(x^4+1)dx]
=1/2[∫(1+1/x²)/(x²+1/x²)dx-∫(1-1/x²)/(x²+1/x²)dx]
=1/2[∫1/(x²+1/x²-2+2)d(x-1/x)-∫1/(x²+1/x²+2-2)d(x+1/x)]
=1/2[∫1/[(x-1/x)²+2]d(x-1/x)-∫1/[(x+1/x)²-2]d(x+1/x)]
=1/2[√2/2arctan[√2/2(x-1/x)]-√2/4ln|(x²-√2x+1)/(x²+√2x+1)|]+C
=√2/4arctan[√2/2(x-1/x)]-√2/8ln|(x²-√2x+1)/(x²+√2x+1)|+C
=1/2∫[(x²+1)-(x²-1)]/(x^4+1)dx
=1/2[∫(x²+1)/(x^4+1)dx-∫(x²-1)/(x^4+1)dx]
=1/2[∫(1+1/x²)/(x²+1/x²)dx-∫(1-1/x²)/(x²+1/x²)dx]
=1/2[∫1/(x²+1/x²-2+2)d(x-1/x)-∫1/(x²+1/x²+2-2)d(x+1/x)]
=1/2[∫1/[(x-1/x)²+2]d(x-1/x)-∫1/[(x+1/x)²-2]d(x+1/x)]
=1/2[√2/2arctan[√2/2(x-1/x)]-√2/4ln|(x²-√2x+1)/(x²+√2x+1)|]+C
=√2/4arctan[√2/2(x-1/x)]-√2/8ln|(x²-√2x+1)/(x²+√2x+1)|+C
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