Question

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

Answer 1

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

Explanation:

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

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

Explanation:

`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`