Question

What is `int (4-x ) / (x^3-6x +4 )`?

Answer 1

`int (4-x)/(x^3-6x+4) dx =`

`=1/3 ln abs(x-2) + (-1-2sqrt(3))/6 ln abs(x+1-sqrt(3))+ (-1+2sqrt(3))/6 ln abs(x+1+sqrt(3)) + C`

Explanation:

Express as a partial fraction decomposition first:

`x^3-6x+4`

`= (x-2)(x^2+2x-2)`

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

`=(x-2)(x+1-sqrt(3))(x+1+sqrt(3))`

Solve:

`(4-x)/(x^3-6x+4)`

`=A/(x-2) + B/(x+1-sqrt(3)) + C/(x+1+sqrt(3))`

`=(A(x^2+2x-2)+B(x-2)(x+1+sqrt(3))+C(x-2)(x+1-sqrt(3)))/(x^3-6x+4)`

`=(A(x^2+2x-2)+B(x^2-(1-sqrt(3))x-2(1+sqrt(3)))+C(x^2-(1+sqrt(3))x-2(1-sqrt(3))))/(x^3-6x+4)`

`=((A+B+C)x^2+(2A-(1-sqrt(3))B-(1+sqrt(3))C)x-2(A+(1+sqrt(3))B+(1-sqrt(3))C))/(x^3-6x+4)`

Equating coefficients, we get the following simulataneous equations:

`(a)color(white)(-)A+B+C = 0`

`(b)color(white)(-)2A-(1-sqrt(3))B-(1+sqrt(3))C = -1`

`(c)color(white)(-)A+(1+sqrt(3))B+(1-sqrt(3))C = -2`

Subtracting `(c)` from `(b)`, we get:

`(d)color(white)(-)A-2B-2C = 1`

Adding twice `(a)` the first equation to `(d)`, we get:

`3A = 1`

So `A = 1/3`

Adding `(b)+(c)` we get:

`(e)color(white)(-)3A+2sqrt(3)B-2sqrt(3)C=-3`

We know `3A = 1`, so this simplifies to:

`2sqrt(3)B-2sqrt(3)C=-4`

Hence:

`(f)color(white)(-)B-C = -2/sqrt(3) = -2sqrt(3)/3`

From `(a)` we get:

`(g)color(white)(-)B+C = -1/3`

Then `(f)+(g)` gives us:

`2B = (-1-2sqrt(3))/3`

and `(g)-(f)` gives us:

`2C = (-1+2sqrt(3))/3`

Hence:

`(4-x)/(x^3-6x+4)`

`=1/(3(x-2)) + (-1-2sqrt(3))/(6(x+1-sqrt(3))) + (-1+2sqrt(3))/(6(x+1+sqrt(3)))`

Then use: `int 1/t dt = ln abs(t) + C` to find:

`int (4-x)/(x^3-6x+4) dx =`

`int 1/(3(x-2)) + (-1-2sqrt(3))/(6(x+1-sqrt(3))) + (-1+2sqrt(3))/(6(x+1+sqrt(3))) dx`

`=1/3 ln abs(x-2) + (-1-2sqrt(3))/6 ln abs(x+1-sqrt(3))+ (-1+2sqrt(3))/6 ln abs(x+1+sqrt(3)) + C`