Question

How do you integrate `(x^2 + 5x - 2)/(x(x^2 - 2))`?

Answer 1

`ln|x| + (5sqrt2)/4 ln |x - sqrt2|/|x + sqrt2| + C`

(Wolfram Alpha gets this too.)

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Always simplify first... You'll get:

`= int cancel(x^2 - 2)/(xcancel((x^2 - 2)))dx + int (5cancel(x))/(cancel(x)(x^2 - 2))dx`

`= int 1/xdx + 5 int 1/(x^2 - 2)dx`

`= ln|x| + 5 int 1/((x + sqrt2)(x - sqrt2))dx`

Now, only the second integral needs partial fractions.

`int 1/((x + sqrt2)(x - sqrt2))dx = int A/(x + sqrt2) + B/(x - sqrt2)dx`

Get common denominators:

`(A(x - sqrt2) + B(x + sqrt2))/cancel((x + sqrt2)(x - sqrt2)) = 1/cancel((x + sqrt2)(x - sqrt2))`

Distribute and get to common polynomial form:

`Ax - Asqrt2 + Bx + Bsqrt2`

`= color(red)((A + B))x + color(green)((-A + B)sqrt2) = color(red)(0)x + color(green)(1)`

Hence, we have the system of equations:

`A + B = 0` `" "" "" "" "" "bb((1))`

`(-A + B)sqrt2 = 1` `" "" "bb((2))`

Solve `(1)` to get `A = -B` and plug into `(2)` to get:

`2B = 1/sqrt2`

`=> B = 1/(2sqrt2) = sqrt2/4`

`=> A = -sqrt2/4`

Therefore, we have:

`int 1/((x + sqrt2)(x - sqrt2))dx`

`= -sqrt2/4 int 1/(x + sqrt2)dx + sqrt2/4 int 1/(x - sqrt2)dx`

`= -sqrt2/4 ln |x + sqrt2| + sqrt2/4 ln |x - sqrt2|`

And thus, the overall integral is:

`int (x^2+5x-2)/(x(x^2-2))dx`

`= ln|x| - (5sqrt2)/4 ln |x + sqrt2| + (5sqrt2)/4 ln |x - sqrt2| + C`

`= color(blue)(ln|x| + (5sqrt2)/4 ln |x - sqrt2|/|x + sqrt2| + C)`