Question

∫(2,0)dx/(x^2-1)^2求不定积分

是定积分

Answer 1

求定积分【0，2】∫dx/(x²-1)²
解：1/(x²-1)²=1/[(x+1)²(x-1)²]=(Ax+B)/(x+1)²-(Cx+D)/(x-1)²
=[(Ax+B)(x-1)²-(Cx+D)(x+1)²]/[(x+1)²(x-1)²]
=[(Ax+B)(x²-2x+1)-(Cx+D)(x²+2x+1)]/[(x+1)²(x-1)²]
=[(Ax³+Bx²-2Ax²-2Bx+Ax+B)-(Cx³+Dx²+2Cx²+2Dx+Cx+D)]/[(x+1)²(x-1)²]
=[(A-C)x³+(B-2A-D-2C)x²+(-2B+A-2D-C)x+B-D]/[(x+1)²(x-1)²]
故得：
A-C=0......................(1)
B-2A-D-2C=0...........(2)
-2B+A-2D-C=0.........(3)
B-D=1......................(4)
由(1)得A=C；代入(3)式得 -2B-2D=0；故得B+D=0........(5)
(4)+(5)得2B=1，故B=1/2；D=-1/2；代入(2)式得：A＋C=1/2.......(6)
(1)+(6)得2A=1/2，故A=1/4，C=1/4；代入原式得：
【0，2】∫dx/(x²-1)²=【0，2】∫[(1/4)x+(1/2)]dx/(x+1)²-【0，2】∫[(1/4)x-(1/2)]dx/(x-1)²
=【0，2】(1/4)∫(x+2)dx/(x+1)²-【0，2】(1/4)∫(x-2)dx/(x-1)²
=【0，2】(1/4)[∫(x+1)dx/(x+1)²+∫dx/(x+1)²-∫(x-1)dx/(x-1)²+∫dx/(x-1)²]
=【0，2】(1/4)[∫dx/(x+1)+∫dx/(x+1)²-∫dx/(x-1)+∫dx/(x-1)²]

=(1/4)[ln(x+1)-1/(x+1)+ln∣x-1∣-1/(x-1)]∣【0，2】
=(1/4){[ln3-(1/3)-1]-[-1+1]}=(1/4)[(ln3)-(4/3)]=(1/4)ln3-(1/3)

Answer 2

∫(2,0)dx/(x^2-1)^2=x/(x^2-1)^2/2,0=2/(2^2-1)^2-0=2/9

追问

第二步怎样来的