Question

高等数学。 请问图中题怎么做？？…！

Answer 1

原式=∫dx/x(x+1)(x²-x+1)=∫[A/x+B/(x+1)+(Cx+D)/(x²-x+1)]dx
解得：A=1，B=-1/3，C=-2/3，D=1/3
所以原式=∫[1/x+(-1/3)/(x+1)+(-2x/3+1/3)/(x²-x+1)]dx
=ln∣x∣+(-1/3)ln∣x+1∣+(-2/3)∫(x-1/2)dx/[(x-1/2)²+3/4]
=ln∣x∣+(-1/3)ln∣x+1∣+(-1/3)∫d(x-1/2)²/[(x-1/2)²+3/4]
=ln∣x∣+(-1/3)ln∣x+1∣+(-1/3)ln[(x-1/2)²+3/4]+C

Answer 2

原积分 I= (1/3)∫[3/x-1/(x+1)-(2x-1)/(x^2-x+1)]dx
= (1/3)[ 3ln|x| - ln|x+1| - ln(x^2-x+1) + C
= (1/3) ln|x^3/(x^3-1)| + C