How do you find ∫x3−x−1(x4−1)xdx using partial fractions?

1 Answer
Cem Sentin
Oct 26, 2017

lnx−14⋅ln(x4−1)+arctanx+C

Explanation:

I decomposed integrand into basic fractions,

x3−x−1x⋅(x4−1)

=x3−x−1x(x+1)(x−1)(x2+1)

=Ax+Bx+1+Cx−1+Dx+Ex2+1

After expanding denominators,

A(x+1)(x−1)(x2+1)+Bx(x−1)(x2+1)+Cx(x+1)(x2+1)+(Dx+E)x(x2−1)=x3−x−1

Set x=−1, 4B=−1orB=−14

Set x=0, −A=−1orA=1

Set x=1, 4C=−1orC=−14

Set x=i, (E+Di)⋅(−2i)=−1−2iorE+Di=1−12⋅i. So D=−12andE=1

Hence,

x3−x−1x⋅(x4−1)

=1x−14⋅1x+1−14⋅(x−1)−12⋅x−2x2+1

Hence,

∫x3−x−1x⋅(x4−1)⋅dx

=∫dxx-14dxx+1-14dxx−1-12*∫(x−2)⋅dxx2+1

=lnx−14⋅ln(x+1)−14⋅ln(x−1)-14*∫2x⋅dxx2+1+dxx2+1

=lnx−14⋅ln(x+1)−14⋅ln(x−1)−14⋅ln(x2+1)+arctanx+C

=lnx−14⋅ln(x4−1)+arctanx+C

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