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"#