Question

How do you integrate `1/(x^3 +4x)` using partial fractions?

Answer 1

The answer is `=1/4ln(|x|)-1/8ln(x^2+4)+C`

Explanation:

Perform the decomposition into partial fractions

`1/(x^3+4x)=1/(x(x^2+4))=A/x+(Bx+C)/(x^2+4)`

`=(A(x^2+4)+x(Bx+C))/(x(x^2+4))`

The denominators are the same, compare the numerators

`1=A(x^2+4)+x(Bx+C)`

Let, `x=0`, `=>`, `1=4A`, `=>`, `A=1/4`

Coefficients of `x^2`

`0=A+B`, `=>`, `B=-A=-1/4`

Coefficients of `x`

`0=C`

Therefore,

`1/(x^3+4x)=(1/4)/x+((-1/4)x)/(x^2+4)`

So,

`int(dx)/(x^3+4x)=int(1/4dx)/x+int((-1/4dx)x)/(x^2+4)`

`=1/4int(dx)/x-1/4int(xdx)/(x^2+4)`

Perform the second integral by substitution

Let `u=x^2+4`, `=>`, `du=2xdx`

So

`1/4int(xdx)/(x^2+4)=1/8int(du)/u=1/8lnu=1/8ln(x^2+4)`

`int(dx)/(x^3+4x)=1/4ln(|x|)-1/8ln(x^2+4)+C`

Answer 2

`1/4*Lnx-1/8*Ln(x^2+4)+C=1/8*Ln(x^2/(x^2+4))+C`

Explanation:

`int dx/(x^3+4x)`

=`int dx/[x*(x^2+4)]`

=`1/4*int (4*dx)/[x*(x^2+4)]`

=`1/4*int [(x^2+4)*dx]/[x(x^2+4)]`-`1/4*int [x^2*dx]/[x(x^2+4)]`

=`1/4*int (dx)/x`-`1/4*int (x*dx)/(x^2+4)`

=`1/4*Lnx-1/8*Ln(x^2+4)+C`

=`1/8*Ln(x^2/(x^2+4))+C`

Answer 3

`1/8ln|x^2/(x^2+4)|+C.`

Explanation:

Here is another way to solve the Problem, but without the

use of Partial Fractions.

Let, `I=int1/(x^3+4x)dx=int1/{x(x^2+4)}dx=intx/{x^2(x^2+4)}dx,` or,

`I=1/2int(2x)/{x^2(x^2+4)}dx.`

Subst. `x^2=t rArr 2xdx=dt.`

`:. I=1/2int1/{t(t+4)}dt=1/2int1/(t^2+4t)dt,`

`=1/2int1/{(t^2+4t+4-4)}dt,`

`=1/2int1/{(t+2)^2-2^2}dt.`

Since, `int1/(y^2-a^2)dy=1/(2a)ln|(y-a)/(y+a)|,`

`I=1/2*1/(2*2)ln|{(t+2)-2}/{(t+2)+2}|,`

`=1/8ln|t/(t+4)|,`

`rArr I=1/8ln|x^2/(x^2+4)|+C.`