|（x^2+1）/((x+1)*(x-2)^2)dx.求不定积分?.求大神指教.

　　先用待定系数法把该有理分式分解为最简分式之和,再行积分.设(x^2+1)/[(x+1)*(x-2)^2] = A/(x+1) + B/(x-2) + C/(x-2)^2,去分母,得x^2+1 = A(x-2)^2 + B(x+1)(x-2) + C(x+1),对比同次项系数,可求得A,B,C（略,留给...