Question

How do you solve this partial fraction? 2x^2-x+4/x^3+4x

Answer 1

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

Explanation:

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

Rather than using partial fractions we can split the function as:

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

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

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

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

Now:

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

Calculate the integrals separately:

`int dx/x = ln abs x +C`

`int (xdx)/(x^2+4) = 1/2 int (d(x^2+4))/(x^2+4) = 1/2ln (x^2+4)+C`

`intdx/(x^2+4) = 1/4 int dx/((x/2)^2+1) = 1/2 int (d(x/2))/((x/2)^2+1) = 1/2 arctan(x/2)+C`

So:

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