高数,解/[(x^2+1)(x^2+x+1)]怎样得出或想出,∫[-x/(x^2+1)+(x+1)/(x^2+x+1)]dx,用待定系数法求不出额.

Question

高数,解/[(x^2+1)(x^2+x+1)]怎样得出或想出,∫[-x/(x^2+1)+(x+1)/(x^2+x+1)]dx
,用待定系数法求不出额.

Answer 1

待定系数法可以啊.
1/[(x²+1)/(x²+x+1)] = (Ax+B)/(x²+1) + (Cx+D)/(x²+x+1)
1 = (Ax+B)(x²+x+1) + (Cx+D)(x²+1)
1 = Ax³+Ax²+Ax + Bx²+Bx+B + Cx³+Cx + Dx²+D
1 = (A+C)x³ + (A+B+D)x² + (A+B+C)x + (B+D)
A+C = 0 ...(i)
A+B+D = 0 ...(ii)
A+B+C = 0 ...(iii)
B+D = 1 ...(iv)
将(i)代入(iii)得出B = 0 ...(v)
将(iv)代入(ii)得出A = -1 ...(vi)
将(v)(vi)代入(iii)、(ii)得出C = D = 1
1/[(x²+1)/(x²+x+1)] = -x/(x²+1) + (x+1)/(x²+x+1)
这个积分的答案,有需要就看看吧,不过你自己应该也会做的.
-∫x/(x²+1) dx + ∫(x+1)/(x²+x+1) dx
令x+1 = E(2x+1)+F
x+1 = 2Ex + E + F
2E = 1 => E = 1/2
E + F = 1 => F = 1/2
= -∫d(x²/2)/(x²+1) + ∫[(1/2)(2x+1)+1/2]/(x²+x+1) dx
= -(1/2)∫d(x²+1)/(x²+1) + (1/2)∫(2x+1)/(x²+x+1) dx + (1/2)∫dx/(x²+x+1)
= -(1/2)ln|x²+1| + (1/2)∫d(x²+x+1)/(x²+x+1) + (1/2)∫d(x+1/2)/[(x+1/2)²+3/4]
= -(1/2)ln|x²+1| + (1/2)ln|x²+x+1| + (1/2)*√(4/3)*arctan[(x+1/2)*√(4/3)] + C
= (1/2)ln|(x²+x+1)/(x²+1)| + (1/√3)arctan[(2x+1)/√3] + C
= (1/2)ln|1+x/(x²+1)| + (1/√3)arctan[(2x+1)/√3] + C