int (x^2-1)/(x^4+x^2+1) dx?

2 Answers
Mar 25, 2018

1/2Ln((x^2-x+1)/(x^2+x+1))+C

Explanation:

int (x^2-1)/(x^4+x^2+1)*dx

=int (1-1/x^2)/(x^2+1+1/x^2)*dx

=int ((1-1/x^2)*dx)/((x+1/x)^2-1)

After using y=x+1/x and dy=(1-1/x^2)*dx transforms, this integral became

int dy/(y^2-1)

=1/2int (2dy)/((y+1)*(y-1))

=1/2int dy/(y-1)-1/2int dy/(y+1)

=1/2Ln(y-1)-1/2Ln(y+1)+C

=1/2Ln((y-1)/(y+1))+C

=1/2Ln((x+1/x-1)/(x+1/x+1))+C

=1/2Ln((x^2-x+1)/(x^2+x+1))+C

Mar 25, 2018

int (x^2-1)/(x^4+x^2+1) dx = 1/2 ln abs(x^2-x+1)-1/2ln abs(x^2+x+1) + C

Explanation:

Note that:

x^4+x^2+1 = (x^2-x+1)(x^2+x+1)

We find:

(x^2-1)/(x^4+x^2+1) = (x-1/2)/(x^2-x+1)-(x+1/2)/(x^2+x+1)

color(white)((x^2-1)/(x^4+x^2+1)) = 1/2((2x-1)/(x^2-x+1))-1/2((2x+1)/(x^2+x+1))

color(white)((x^2-1)/(x^4+x^2+1)) = 1/2((d/(dx)(x^2-x+1))/(x^2-x+1))-1/2((d/(dx)(x^2+x+1))/(x^2+x+1))

So:

int (x^2-1)/(x^4+x^2+1) dx = int 1/2((2x-1)/(x^2-x+1))-1/2((2x+1)/(x^2+x+1)) dx

color(white)(int (x^2-1)/(x^4+x^2+1) dx) = 1/2 ln abs(x^2-x+1)-1/2ln abs(x^2+x+1) + C