大学高数,特殊函数的积分.

Question

大学高数,特殊函数的积分.
 

Answer 1

∫1/(x²+1)(x²+x+1)
=∫[(-x)/(x²+1)+(x+1)/(x²+x+1)]
=∫[(-x)/(x²+1)+(x+1)/(x²+x+1)]
=(-1/2)ln(x²+1)+∫(x+1)/(x²+x+1)

∫(x+1)/(x²+x+1)
令x+1/2＝t,则：dx＝dt.
∴∫[(x+1)/（x²+x+1)]dx
＝∫{[(x+1/2)+1/2]/[(x²+x+1/4)+3/4]}dx
＝∫{[(x+1/2)+1/2]/[(x+1/2)²+3/4]}d(x+1/2)
＝∫[(t+1/2)/(t²+3/4)]dt
＝∫{t/(t²+3/4)]dt+(1/2)∫[1/(t²+3/4)]dt
＝(1/2)∫[1/(t²+3/4)]d(t²+3/4)+(1/2)×(4/3)∫[1/[(2t/√3)²+1]}dt
＝(1/2)ln(t²+3/4)+(2/3)×(√3/2)∫{1/[(2t/√3)²+1]}d(2t/√3)
＝(1/2)ln[(x+1/2)²+3/4]+(√3/3)arctan(2t/√3)+C
＝(1/2)ln(x²+x+1)+(√3/3)arctan[2(x+1/2)/√3]+C
＝(1/2)ln(x²+x+1)+(√3/3)arctan[(2x+1)/√3]+C.

∫1/(x²+1)(x²+x+1)
=-(1/2)ln(x²+1)+(1/2)ln(x²+x+1)+(√3/3)arctan[(2x+1)/√3]+C.