求∫ln(X^2+4)dx不定積分
求∫ln(X^2+4)dx不定積分
數(shù)學(xué)人氣:645 ℃時(shí)間:2020-05-20 19:16:45
優(yōu)質(zhì)解答
先分布積分,∫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...∫[(3x^3+2x^2+1)/(x+2)(x-2)(x-1)]dx求不定積分(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)]恒等式兩邊對應(yīng)系數(shù)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
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