Question

一道定积分题（1）求值（2）求f（x）=（里面那堆东西）的原函数

Answer 1

∫(2-x)dx/(x^2-x+1)=∫(-1/2)(2x-1)dx/(x^2-x+1)+∫(3/2)dx/[(x-1/2)^2+3/4]=(-1/2)ln|x^2-x+1|+ √3∫d[(2x-1)/√3]/[(2x-1)^2/3+1]=(-1/2)ln|x^2-x+1|+√3arctan[(2x-1)/√3] +C∫[-1,1] (2-x)dx/(x^2-x+1)=(1/2)...