How do you integrate #1 / (x(x^4+1))# using partial fractions?

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Answer 1 · 2017-12-08T08:02:49

The answer is #=ln(|x|)-1/4ln(x^4+1)+C#

Perform the decomposition into partial fractions

#1/(x(x^4+1))=A/x+(Bx^3+C)/(x^4+1)#

#=(A(x^4+1)+x(Bx^3+C))/(x(x^4+1))#

The denominators are the same, compare the numerators

#1=A(x^4+1)+x(Bx^3+C)#

Let #x=0#, #=>#, #1=A#

Coefficients of #x^4#

#0=A+B#, #=>#, #B=-A=-1#

Coefficients of #x#

#0=C#

Therefore,

#1/(x(x^4+1))=1/x+(-x^3)/(x^4+1)#

So,

#int(dx)/(x(x^4+1))=intdx/x-int(x^3dx)/(x^4+1)#

#=ln(|x|)-int(x^3dx)/(x^4+1)#

Perform the second part by substitution

Let #u=x^4+1#, #=>#, #du=4x^3dx#

So,

#int(x^3dx)/(x^4+1)=1/4int(du)/u#

#=1/4lnu#

#=1/4ln(x^4+1)#

Finally,

#int(dx)/(x(x^4+1))=ln(|x|)-1/4ln(x^4+1)+C#

Answer 2 · 2017-12-08T12:06:21

#int dx/[x*(x^4+1)]=Lnx-1/4Ln(x^4+1)+C#

#int dx/[x*(x^4+1)]#

=#int ((x^4+1)*dx)/[x(x^4+1)]#-#int (x^4*dx)/[x(x^4+1)]#

=#int (dx)/x#-#int (x^3*dx)/(x^4+1)#

=#Lnx-1/4int (4x^3*dx)/(x^4+1)#

=#Lnx-1/4Ln(x^4+1)+C#