求∫ln(X^2+4)dx不定积分

先分布积分,∫ln(x^2+4)dx=xln(x^2+4)-∫xdln(x^2+4)=xln(x^2+4)-∫[2x^2/(x^2+4)]dx=xln(x^2+4)-2∫[x^2/(x^2+4)]dx=xln(x^2+4)-2∫[(x^2+4-4)/(x^2+4)]dx=xln(x^2+4)-2∫[1-4/(x^2+4)]dx=xln(x^2+4)-2x+2∫[4/(x^2+4)]dx=xln(x^2+4)-2x+2∫[1/(x^2/4+1)]dx=xln(x^2+4)-2x+4∫{1/[(x/2)^2+1)]}d(x/2)=xln(x^2+4)-2x+4arctan(x/2)+C 再问： ∫[（3x^3+2x^2+1)/(x+2)(x-2)(x-1)]dx求不定积分 再答： (3x^3+2x^2+1)/(x+2)(x-2)(x-1)=(3x^3+2x^2+1)/(x+2)(x-2)(x-1)=(3x^3+2x^2+1)/(x^3-x^2-4x+4) =[3(x^3-x^2-4x+4)+5x^2+12x-11]/(x^3-x^2-4x+4) =3+(5x^2+12x-11)/(x^3-x^2-4x+4) 令(5x^2+12x-11)/[(x+2)(x-2)(x-1)]=A/(x+2)+B/(x-2)+C/(x-1) =[(A+B+C)x^2+(B-3A)x+2A-2B-4C]/[(x+2)(x-2)(x-1)] 恒等式两边对应系数相等，A+B+C=5，B-3A=12，2A-2B-4C=-11， 解出A，B，C， ∫[（3x^3+2x^2+1)/(x+2)(x-2)(x-1)]dx=∫[3+A/(x+2)+B/(x-2)+C/(x-1)]dx =3x+Aln|x+2|+Bln|x-2|+Cln|x-1|+C1