微积分, 求不定积分 ∫1/(1+x^4)dx
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答：
我曾经答过一样的题.
原式
=∫(x^2+1)/[2(x^4+1)]dx-∫(x^2-1)/[2(x^4+1)]dx
=1/2∫(1+1/x^2)/(x^2+1/x^2)dx-1/2∫(1-1/x^2)/(x^2+1/x^2)dx
=1/2∫d(x-1/x)/[(x-1/x)^2+2]-1/2∫d(x+1/x)/[(x+1/x)^2-2]
=1/4∫d(x-1/x)/[(x-1/x)^2/2+1]-1/2∫d(x+1/x)/[(x+1/x+√2)(x+1/x-√2)]
=√2/4*arctan[(x-1/x)/√2]-1/4∫d(x+1/x)/(x+1/x+√2)-1/4∫d(x+1/x)/(x+1/x-√2)
=√2/4*arctan[(x-1/x)/√2]-1/4*ln|x+1/x+√2|-1/4*ln|x+1/x-√2| +C
我曾经答过一样的题.
原式
=∫(x^2+1)/[2(x^4+1)]dx-∫(x^2-1)/[2(x^4+1)]dx
=1/2∫(1+1/x^2)/(x^2+1/x^2)dx-1/2∫(1-1/x^2)/(x^2+1/x^2)dx
=1/2∫d(x-1/x)/[(x-1/x)^2+2]-1/2∫d(x+1/x)/[(x+1/x)^2-2]
=1/4∫d(x-1/x)/[(x-1/x)^2/2+1]-1/2∫d(x+1/x)/[(x+1/x+√2)(x+1/x-√2)]
=√2/4*arctan[(x-1/x)/√2]-1/4∫d(x+1/x)/(x+1/x+√2)-1/4∫d(x+1/x)/(x+1/x-√2)
=√2/4*arctan[(x-1/x)/√2]-1/4*ln|x+1/x+√2|-1/4*ln|x+1/x-√2| +C
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