Question

Question #cf607

Answer 1

`int (x^3+x+2/x^2-4) = 1/4x^4 + 1/2x^2 -2x^-1 + 4x + c`
where `c in RR`

Explanation:

since this seems like a simple polynomial (oh well there is that rational part) we can do it part by part.

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

Let's think how this works... It's easier to think about integration in terms of derivitives

for example we know that `f'(x^4)= 4x^3`

So if we wanted to get rid of the `4`, we can multiply it by `1/4`, so they both cross out., thus `int (x^3)= 1/4x^4 + c`

same idea for `int (x)` thus `int(x)= 1/2x^2 + c`

we can use the same idea for `2/x^2`, if we turn it into this format `2x^-2`

(note that I took out the `2` because it's a coefficient and it doesn't affect the integration.)

!Attention! the `-2` does not become `-3`, remember that when we derive a polynomial, we SUBSTRACT one from the power, so when we integrate we ADD one to the power. so `-2` becomes `-1`
`2 f'(x^-1) = -x^-2`
multiply by -1 to get rid of the negative sign

therefore: `2int(x^-2) = -2x^-1 +c`

finally, let's think when do we end up with a single number in a derivative? when we have smth like this `f'(x) = 1`
so `f'(4x)= 4`
thus `int (4) = 4x + c`

put all terms together (just put one c, or C it just represents a constant)
`int (x^3+x+2/x^2-4) = 1/4x^4 + 1/2x^2 -2x^-1 + 4x + c`
where `c in RR`