求函数 ln（x+1）/（x4+x2+1） 从0到100（积分上下限）上定积分的值,高手帮下忙,不要只语言叙述,最好能贴张图~这题困扰我好几天了,.ln（x+1）整个是分子，我写的不严谨不好意思。

求函数 ln（x+1）/（x4+x2+1） 从0到100（积分上下限）上定积分的值,高手帮下忙,不要只语言叙述,最好能贴张图~这题困扰我好几天了,.ln（x+1）整个是分子，我写的不严谨不好意思。
求函数 ln（x+1）/（x4+x2+1） 从0到100（积分上下限）上定积分的值,高手帮下忙,
不要只语言叙述,最好能贴张图~这题困扰我好几天了,.
ln（x+1）整个是分子，我写的不严谨不好意思。

求函数 ln（x+1）/（x4+x2+1） 从0到100（积分上下限）上定积分的值,高手帮下忙,不要只语言叙述,最好能贴张图~这题困扰我好几天了,.ln（x+1）整个是分子，我写的不严谨不好意思。
[0,100]∫ln[(x+1)/(x⁴+x²+1)]dx=[0,100][∫ln(x+1)dx-∫ln(x⁴+x²+1)dx].(1)
为简化书写过程我先把两个不定积分求出来：
∫ln(x+1)dx=∫ln(x+1)d(x+1)=(x+1)ln(x+1)-x.(2)
∫ln(x⁴+x²+1)dx=xln(x⁴+x²+1)-∫[x(4x³+2x)/(x⁴+x²+1)]dx
=xln(x⁴+x²+1)-∫[4-(2x²+4)/(x⁴+x²+1)]dx
=xln(x⁴+x²+1)-4x+2∫[(x²+2)/(x⁴+x²+1)]dx.(3)
其中∫[(x²+2)/(x⁴+x²+1)]dx=∫[(x²+1)/(x⁴+x²+1)]dx+∫[1/(x⁴+x²+1)]dx
=(1/√3)arctan[(x²-1)/(√3)x]+(1/4)ln[(x²+x+1)/(x²-x+1)]+(1/2√3)arctan[(x²-1)/(√3)x]
=(3/2√3)arctan[(x²-1)/(√3)x]+(1/4)ln[(x²+x+1)/(x²-x+1)],代入(3)式得：
∫ln(x⁴+x²+1)dx=xln(x⁴+x²+1)-4x+(3/√3)arctan[(x²-1)/(√3)x]+(1/2)ln[(x²+x+1)/(x²-x+1)].(4)
再将(2)和(4)代入(1)式即得：
[0,100]∫ln[(x+1)/(x⁴+x²+1)]dx
=[0,100]{(x+1)ln(x+1)-xln(x⁴+x²+1)+4x-(3/√3)arctan[(x²-1)/(√3)x]-(1/2)ln[(x²+x+1)/(x²-x+1)]}
=(101)ln(101)-100ln(100010001)+400-(3/√3)arctan(9999/100√3)
-(1/2)ln(100010001/99989999)-3π/(2√3)
你这题够烦人的!中间有两个积分把过程省去了,若写下来更不得了!

1/12*i*3^(1/2)*dilog(202/(1+i*3^(1/2)))+1/12*i*3^(1/2)*dilog(202/(3+i*3^(1/2)))+1/8*log(1-i*3^(1/2))*log(3)+1/8*log(1-i*3^(1/2))*log(7753)+1/8*log(1+i*3^(1/2))*log(7753)-1/4*log(1-i*3^(1/2))*log(-99-1...

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1/12*i*3^(1/2)*dilog(202/(1+i*3^(1/2)))+1/12*i*3^(1/2)*dilog(202/(3+i*3^(1/2)))+1/8*log(1-i*3^(1/2))*log(3)+1/8*log(1-i*3^(1/2))*log(7753)+1/8*log(1+i*3^(1/2))*log(7753)-1/4*log(1-i*3^(1/2))*log(-99-101*i*3^(1/2))+1/12*i*3^(1/2)*log(3-i*3^(1/2))*log(151)-1/4*log(2)*log(7753)-1/4*log(2)*log(3)+1/4*log(3-i*3^(1/2))*log(-297-101*i*3^(1/2))-1/4*log(3+i*3^(1/2))*log(151)-1/4*log(202)*log(-297-101*i*3^(1/2))+1/12*i*3^(1/2)*dilog(1+202/(-1+i*3^(1/2)))+1/8*log(1+i*3^(1/2))*log(3)+1/8*log(1-i*3^(1/2))*log(7)+1/12*i*3^(1/2)*log(2)^2+1/12*i*3^(1/2)*log(3-i*3^(1/2))^2+1/12*i*3^(1/2)*log(1-i*3^(1/2))^2+1/8*log(1+i*3^(1/2))*log(7)+1/4*dilog(2/(1+i*3^(1/2)))+1/4*log(2)^2-1/12*i*3^(1/2)*log(1-i*3^(1/2))*log(2)-1/6*3^(1/2)*atan(1/9*3^(1/2))*log(2)-1/4*dilog(202/(1+i*3^(1/2)))+1/4*dilog(202/(3+i*3^(1/2)))+1/12*i*3^(1/2)*dilog(1+202/(-3+i*3^(1/2)))+1/4*dilog(1+202/(-1+i*3^(1/2)))-1/4*dilog(1+202/(-3+i*3^(1/2)))-1/12*i*3^(1/2)*dilog(2/(3+i*3^(1/2)))+1/24*i*3^(1/2)*log(1-i*3^(1/2))*log(7)-1/4*i*atan(1/9*3^(1/2))*log(3+i*3^(1/2))-1/4*dilog(2/(3+i*3^(1/2)))+1/24*i*3^(1/2)*log(1-i*3^(1/2))*log(7753)-1/12*i*3^(1/2)*log(3+i*3^(1/2))*log(151)-1/4*log(2)*log(6)+1/4*log(2)*log(3-i*3^(1/2))-1/12*3^(1/2)*atan(7/3*3^(1/2))*log(1+i*3^(1/2))-1/4*log(3-i*3^(1/2))*log(151)+1/6*3^(1/2)*atan(7/3*3^(1/2))*log(2)+1/12*3^(1/2)*atan(1/9*3^(1/2))*log(3+i*3^(1/2))-1/4*log(202)*log(2)-1/4*i*atan(7/3*3^(1/2))*log(1+i*3^(1/2))+1/4*log(202)*log(-99-101*i*3^(1/2))+1/4*log(202)*log(6)-1/12*i*3^(1/2)*log(202)*log(2)+1/12*i*3^(1/2)*log(202)*log(-99-101*i*3^(1/2))+1/24*i*3^(1/2)*log(1-i*3^(1/2))*log(3)+1/4*i*atan(1/9*3^(1/2))*log(3-i*3^(1/2))+1/4*i*atan(7/3*3^(1/2))*log(1-i*3^(1/2))+1/12*i*3^(1/2)*log(2)*log(6)-1/12*i*3^(1/2)*log(2)*log(3-i*3^(1/2))+1/4*log(1-i*3^(1/2))^2-1/4*log(3-i*3^(1/2))^2-1/12*i*3^(1/2)*log(202)*log(6)+1/12*i*3^(1/2)*log(202)*log(-297-101*i*3^(1/2))-1/12*i*3^(1/2)*log(3-i*3^(1/2))*log(-297-101*i*3^(1/2))-1/12*3^(1/2)*atan(7/3*3^(1/2))*log(1-i*3^(1/2))-1/12*i*3^(1/2)*dilog(2/(1+i*3^(1/2)))+1/2*log(2)*log(151)-1/24*i*3^(1/2)*log(1+i*3^(1/2))*log(7)+1/12*3^(1/2)*atan(1/9*3^(1/2))*log(3-i*3^(1/2))-1/6*3^(1/2)*atan(799/9*3^(1/2))*log(2)+1/12*3^(1/2)*atan(799/9*3^(1/2))*log(3-i*3^(1/2))-1/4*i*atan(269*3^(1/2))*log(1-i*3^(1/2))+1/12*3^(1/2)*atan(799/9*3^(1/2))*log(3+i*3^(1/2))+1/4*i*atan(269*3^(1/2))*log(1+i*3^(1/2))-1/6*3^(1/2)*atan(269*3^(1/2))*log(2)+1/12*3^(1/2)*atan(269*3^(1/2))*log(1+i*3^(1/2))-1/4*i*atan(799/9*3^(1/2))*log(3+i*3^(1/2))+1/12*3^(1/2)*atan(269*3^(1/2))*log(1-i*3^(1/2))+1/4*i*atan(799/9*3^(1/2))*log(3-i*3^(1/2))-1/24*i*3^(1/2)*log(1+i*3^(1/2))*log(3)-1/12*i*3^(1/2)*log(1-i*3^(1/2))*log(-99-101*i*3^(1/2))-1/12*i*3^(1/2)*dilog(1+2/(-1+i*3^(1/2)))-1/12*i*3^(1/2)*dilog(1+2/(-3+i*3^(1/2)))-1/24*i*3^(1/2)*log(1+i*3^(1/2))*log(7753)+1/24*i*3^(1/2)*log(1+i*3^(1/2))*log(13)-1/24*i*3^(1/2)*log(1-i*3^(1/2))*log(13)-1/4*log(1-i*3^(1/2))*log(2)-1/4*dilog(1+2/(-1+i*3^(1/2)))-1/4*log(2)*log(7)+1/4*log(2)*log(13)-1/8*log(1+i*3^(1/2))*log(13)-1/8*log(1-i*3^(1/2))*log(13)+1/4*dilog(1+2/(-3+i*3^(1/2)))

上面是用matlab自动得出的结果，不应该错。这道题的思路是：用分布积分，令u=ln(1+x) dv为其余部分，其余部分是有理函数，能得出原函数，而u的导数是1/(1+x),简单了。以后多次分布积分，麻烦一点，不怕费时间您就慢慢做吧。

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