How do you evaluate the integral $int x^4/(x^2-1)dx$ ?

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Answer 1 · 2018-03-04T20:29:54

$int x^4/(x^2-1)dx =1/3x^3+x+1/2lnabs((x-1)/(x+1))+"c"$

For the integrand $x^4/(x^2-1)$ , we perform a long division to get it into an integrable form.

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

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

We now perform a partial fraction decomposition on the 2nd term

$1/((x+1)(x-1))=((x+1)-(x-1))/(2(x+1)(x-1))=$

$(x+1)/(2(x+1)(x-1))-(x-1)/(2(x+1)(x-1))=1/(2(x-1))-1/(2(x+1))$

So

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

and we can now integrate:

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

$=1/3x^3+x+1/2lnabs(x-1)-1/2lnabs(x+1)+"c"$

$=1/3x^3+x+1/2lnabs((x-1)/(x+1))+"c"$

How do you evaluate the integral int x^4/(x^2-1)dxint x^4/(x^2-1)dx?