Question

∫x²/(x^4+x²+1)dx怎么做啊，求助

Answer 1

令A=∫x^2/(x^4+x^2+1)dx，B=∫1/(x^4+x^2+1)dx
A+B=∫(x^2+1)/(x^4+x^2+1)dx
=∫(1+1/x^2)/(x^2+1+1/x^2)dx
=∫d(x-1/x)/[(x-1/x)^2+3]
=(1/√3)*arctan[(x-1/x)/√3]+C1，其中C1是任意常数
A-B=∫(x^2-1)/(x^4+x^2+1)dx
=∫(1-1/x^2)/(x^2+1+1/x^2)dx
=∫d(x+1/x)/[(x+1/x)^2-1]
=∫d(x+1/x)/(x+1/x+1)(x+1/x-1)
=(1/2)*∫[1/(x+1/x-1)-1/(x+1/x+1)]d(x+1/x)
=(1/2)*[ln|x+1/x-1|-ln|x+1/x+1|]+C2，其中C2是任意常数
所以A=(1/2)*[(A+B)+(A-B)]
=(1/2)*{(1/√3)*arctan[(x-1/x)/√3]+(1/2)*[ln|x+1/x-1|-ln|x+1/x+1|]+C
=(√3/6)*arctan[(x-1/x)/√3]+(1/4)*ln|x+1/x-1|-(1/4)*ln|x+1/x+1|+C，其中C是任意常数