作业帮求1/(x4+1)的不定积分
来源：学生作业帮 编辑：作业帮 分类：数学作业 时间：2024/07/14 09:10:21
求1/(x4+1)的不定积分
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
=(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
求1/(x的四次方+1) 的的不定积分 即 S 1/( x4 + 1 ) dx
高数求不定积分,∫x4/(1+x2)dx
求不定积分 arcsinx的不定积分 e^√x+1的不定积分 (x-1)lnx的不定积分
求不定积分1.厂珈玛的意思.厂(5x4－2x＋1)dx2丶厂cos(2－3x)dx3丶厂(2x＋3)10dx求不定积分
已知x2+3x+1=0,求x4+1/x4的值.
已知x2+x+1=0,求x4+1/x4的值
如果x平方-3x+1=0,求x4次方+x4次方分之一的值.
已知x2+x+1=0 求x4+1/x4
求x-1分之1的不定积分
求1/(1+t^4)的不定积分
1/(1 cost)的不定积分怎么求
求1/(e^x-1)的不定积分