For the integrand x^4/(x^2-1)x4x2−1, we perform a long division to get it into an integrable form.
x^4/(x^2-1)=(x^4-1+1)/(x^2-1)=((x^2-1)(x^2+1))/(x^2-1)+1/(x^2-1)=x4x2−1=x4−1+1x2−1=(x2−1)(x2+1)x2−1+1x2−1=
=x^2+1+1/((x+1)(x-1))=x2+1+1(x+1)(x−1)
We now perform a partial fraction decomposition on the 2nd term
1/((x+1)(x-1))=((x+1)-(x-1))/(2(x+1)(x-1))=1(x+1)(x−1)=(x+1)−(x−1)2(x+1)(x−1)=
(x+1)/(2(x+1)(x-1))-(x-1)/(2(x+1)(x-1))=1/(2(x-1))-1/(2(x+1))x+12(x+1)(x−1)−x−12(x+1)(x−1)=12(x−1)−12(x+1)
So
x^4/(x^2-1)=x^2+1+1/(2(x-1))-1/(2(x+1))x4x2−1=x2+1+12(x−1)−12(x+1)
and we can now integrate:
intx^4/(x^2-1)dx=intx^2+1+1/(2(x-1))-1/(2(x+1))dx∫x4x2−1dx=∫x2+1+12(x−1)−12(x+1)dx
=1/3x^3+x+1/2lnabs(x-1)-1/2lnabs(x+1)+"c"=13x3+x+12ln|x−1|−12ln|x+1|+c
=1/3x^3+x+1/2lnabs((x-1)/(x+1))+"c"=13x3+x+12ln∣∣∣x−1x+1∣∣∣+c
