∫[1/(x-2)(x-1)²]dx＝？

∫ 1/(x²+x+1)² dx= 4/(3√3)*arctan[(2x+1)/√3] + (2x+1)/[3(x²+x+1)] + C.C为积分常数.解答过程如下:∫1/(x²+x+1)² dx= ∫1/[(x+1/2)²+3/4]² dx 令x+1/2=√3/2*tanθ,...

∫(1+x²)²dx=∫(1+2x²+x^4)dx=∫dx+2∫x²dx+∫x^4dx=x+(2/3)x³+(1/5)x^5+c ∫(1-x²)²dx=∫(1-2x²+x^4)dx=∫dx-2∫x²dx+∫x^4dx=x-(2/3)x³+(1/5)x^5+c

∫x(1-x^2)^(1/2) dx=0.5 *∫(1-x^2)^(1/2) d(x^2)= -0.5 *∫(1-x^2)^(1/2) d(1-x^2)= -0.5 *2/3 *(1-x^2)^(3/2) +C= -1/3 *(1-x^2)^(3/2) +C,C为常数