How do you find int 2/((x^4-1)x) dx2(x41)xdx using partial fractions?

1 Answer
Oct 30, 2015

I=-2ln|x|+1/2ln|x-1|+1/2ln|x+1|+1/2ln|x^2+1|+CI=2ln|x|+12ln|x1|+12ln|x+1|+12lnx2+1+C

Explanation:

2/((x^4-1)x)=2/((x^2-1)(x^2+1)x)=2/((x-1)(x+1)(x^2+1)x)2(x41)x=2(x21)(x2+1)x=2(x1)(x+1)(x2+1)x

2/((x-1)(x+1)(x^2+1)x)=A/x+B/(x-1)+C/(x+1)+(Dx+E)/(x^2+1)=2(x1)(x+1)(x2+1)x=Ax+Bx1+Cx+1+Dx+Ex2+1=

=(A(x-1)(x+1)(x^2+1))/(x(x-1)(x+1)(x^2+1))+(Bx(x+1)(x^2+1))/(x(x-1)(x+1)(x^2+1))+=A(x1)(x+1)(x2+1)x(x1)(x+1)(x2+1)+Bx(x+1)(x2+1)x(x1)(x+1)(x2+1)+

+(Cx(x-1)(x^2+1))/(x(x-1)(x+1)(x^2+1))+((Dx+E)x(x-1)(x+1))/(x(x-1)(x+1)(x^2+1))=+Cx(x1)(x2+1)x(x1)(x+1)(x2+1)+(Dx+E)x(x1)(x+1)x(x1)(x+1)(x2+1)=

=(A(x^4-1))/(x(x-1)(x+1)(x^2+1))+(Bx(x^3+x^2+x+1))/(x(x-1)(x+1)(x^2+1))+=A(x41)x(x1)(x+1)(x2+1)+Bx(x3+x2+x+1)x(x1)(x+1)(x2+1)+

+(Cx(x^3-x^2+x-1))/(x(x-1)(x+1)(x^2+1))+((Dx+E)(x^3-x))/(x(x-1)(x+1)(x^2+1))=+Cx(x3x2+x1)x(x1)(x+1)(x2+1)+(Dx+E)(x3x)x(x1)(x+1)(x2+1)=

=(Ax^4-A)/(x(x-1)(x+1)(x^2+1))+(Bx^4+Bx^3+Bx^2+Bx)/(x(x-1)(x+1)(x^2+1))+=Ax4Ax(x1)(x+1)(x2+1)+Bx4+Bx3+Bx2+Bxx(x1)(x+1)(x2+1)+

+(Cx^4-Cx^3+Cx^2-Cx)/(x(x-1)(x+1)(x^2+1))+(Dx^4+Ex^3-Dx^2-Ex)/(x(x-1)(x+1)(x^2+1))=+Cx4Cx3+Cx2Cxx(x1)(x+1)(x2+1)+Dx4+Ex3Dx2Exx(x1)(x+1)(x2+1)=

=(x^4(A+B+C+D)+x^3(B-C+E))/(x(x-1)(x+1)(x^2+1))+=x4(A+B+C+D)+x3(BC+E)x(x1)(x+1)(x2+1)+

+(x^2(B+C-D)+x(B-C-E)+(-A))/(x(x-1)(x+1)(x^2+1))+x2(B+CD)+x(BCE)+(A)x(x1)(x+1)(x2+1)

A+B+C+D=0A+B+C+D=0
B-C+E=0BC+E=0
B+C-D=0B+CD=0
B-C-E=0BCE=0
-A=2 => A=-2A=2A=2

[II-IV] => 2E=0 => E=0[IIIV]2E=0E=0

B+C+D=2B+C+D=2
B-C=0 => B=CBC=0B=C
B+C-D=0B+CD=0

2C+D=22C+D=2
2C-D=02CD=0

4C=2 => C=1/2 => B=1/24C=2C=12B=12

D=2C => D=1D=2CD=1

I=int 2/((x^4-1)x)dx= -2int dx/x +1/2int dx/(x-1) + 1/2int dx/(x+1) + int (xdx)/(x^2+1)I=2(x41)xdx=2dxx+12dxx1+12dxx+1+xdxx2+1

I=-2ln|x|+1/2ln|x-1|+1/2ln|x+1|+1/2ln|x^2+1|+CI=2ln|x|+12ln|x1|+12ln|x+1|+12lnx2+1+C

Note:

int (xdx)/(x^2+1) = int (1/2(2xdx))/(x^2+1) = 1/2int (d(x^2+1))/(x^2+1) = 1/2ln|x^2+1|xdxx2+1=12(2xdx)x2+1=12d(x2+1)x2+1=12lnx2+1