y=1/(x^4 + x^3 + x^2 + x + 1)用matlab怎么求（1,2）区间的定积分?要数值解

y=1/(x^4 + x^3 + x^2 + x + 1)用matlab怎么求（1,2）区间的定积分?要数值解
y=1/(x^4 + x^3 + x^2 + x + 1)用matlab怎么求（1,2）区间的定积分?要数值解

y=1/(x^4 + x^3 + x^2 + x + 1)用matlab怎么求（1,2）区间的定积分?要数值解
用数值积分函数,可以直接得到结果：
>> f=@(x)1./(x.^4+x.^3+x.^2+x+1);
>> y=quad(f,1,2)
y =
0.0888

>> syms x
>> int(1/(x^4 + x^3 + x^2 + x + 1),1,2)

ans =

-1/10*(5^(1/2)*log(6-5^(1/2))*(10+2*5^(1/2))^(1/2)*(10-2*5^(1/2))^(1/2)-10*atan(9/(10+2*5^(1/2))^(1/2)-1/(10+2*5^(1/2))^...

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>> syms x
>> int(1/(x^4 + x^3 + x^2 + x + 1),1,2)

ans =

-1/10*(5^(1/2)*log(6-5^(1/2))*(10+2*5^(1/2))^(1/2)*(10-2*5^(1/2))^(1/2)-10*atan(9/(10+2*5^(1/2))^(1/2)-1/(10+2*5^(1/2))^(1/2)*5^(1/2))*(10-2*5^(1/2))^(1/2)+2*atan(9/(10+2*5^(1/2))^(1/2)-1/(10+2*5^(1/2))^(1/2)*5^(1/2))*5^(1/2)*(10-2*5^(1/2))^(1/2)-5^(1/2)*log(6+5^(1/2))*(10+2*5^(1/2))^(1/2)*(10-2*5^(1/2))^(1/2)-10*atan(9/(10-2*5^(1/2))^(1/2)+1/(10-2*5^(1/2))^(1/2)*5^(1/2))*(10+2*5^(1/2))^(1/2)-2*atan(9/(10-2*5^(1/2))^(1/2)+1/(10-2*5^(1/2))^(1/2)*5^(1/2))*5^(1/2)*(10+2*5^(1/2))^(1/2)-5^(1/2)*log(5-5^(1/2))*(10+2*5^(1/2))^(1/2)*(10-2*5^(1/2))^(1/2)+10*atan(5/(10+2*5^(1/2))^(1/2)-1/(10+2*5^(1/2))^(1/2)*5^(1/2))*(10-2*5^(1/2))^(1/2)-2*atan(5/(10+2*5^(1/2))^(1/2)-1/(10+2*5^(1/2))^(1/2)*5^(1/2))*5^(1/2)*(10-2*5^(1/2))^(1/2)+5^(1/2)*log(5+5^(1/2))*(10+2*5^(1/2))^(1/2)*(10-2*5^(1/2))^(1/2)+10*atan(5/(10-2*5^(1/2))^(1/2)+1/(10-2*5^(1/2))^(1/2)*5^(1/2))*(10+2*5^(1/2))^(1/2)+2*atan(5/(10-2*5^(1/2))^(1/2)+1/(10-2*5^(1/2))^(1/2)*5^(1/2))*5^(1/2)*(10+2*5^(1/2))^(1/2))/(10+2*5^(1/2))^(1/2)/(10-2*5^(1/2))^(1/2)

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