Question

求1/(x4+1)的不定积分

解答详细点

Answer 1

设1\(x^4+1)
=(ax+b)\[x^2+2^(1\2)x+1]+(cx+d)\[x^2-2^(1\2)x+1]
则a+c=0 b+d+2^(1\2)(c-a)=0
a+c+2^(1\2)(d-b)=0 b+d=1
a=2^(1\2)\4 c=-2^(1\2)\4 b=d=1\2
∴1\(x^4+1)=[2^(1\2)\8]*{[2x+2*2^(1\2)]\[x^2+2^(1\2)x+1]
-[2x-2*2^(1\2)]\[x^2-2^(1\2)x+1]}
=[2^(1\2)\8]*{[2x+2^(1\2)]\[x^2+2^(1\2)x+1]
-[2x-2^(1\2)]\[x^2-2^(1\2)x+1]}+
1\4{[x+2^(1\2)\2]^2+1\2}+1\4{[x-2^(1\2)\2]^2+1\2}
∴∫dx\(x^4+1)
=[2^(1\2)\8]*In{[x^2+2^(1\2)x+1]\[x^2-2^(1\2)x+1]}
+[2^(1\2)\4]*{arctan[2^(1\2)x+1]+arctan[2^(1\2)x-1]}
+C
=[2^(1\2)\8]*In{[x^2+2^(1\2)x+1]\[x^2-2^(1\2)x+1]}
+[2^(1\2)\4]*arctan[2^(1\2)x\(1-x^2)]+C

Answer 2

可使1/(x4+1)等价于1/2（（x2+1）/(X4+1)-(x2—1)/（x4+1））然后分别同除于x2可求得上述两个积分 （楼主有点麻烦）