find f(x)
f'(x) = (x^2-1)/x^(-1)
What exactly was wrong with my last response?
i just didn't understand how you got to that point. And i havn't learned integrals yet. I was hoping someone else could give me a different perspective as to how to answer it.
im sorry. i think it was because i wrote it wrong.
i meant
f'(x) = (x^2-1)/x
Note that f'(x) can be rewritten algebraically a f'(x)= x -(1/x).
Now, take the integral (anti-drivitive)
of f'(x) with respect to 'x' to obtain
f(x).
So,
Int(x-(1/x),x)=((x^(2))/2)-ln(abs(x))+C
To understand this you must first look at the rules of integrals.
By the way in some computer programs if you enter this, for some reason they don't like the ln(abs(x)) even though this is correct so just make it ln(x).
To find f(x), we can integrate the given derivative expression. In this case, let's integrate the expression f'(x) = (x^2 - 1) / x^(-1) with respect to x.
To integrate the expression, we use the power rule of integration, which states that the integral of x^n with respect to x is (x^(n+1))/(n+1), except when n equals -1. In the case of n = -1, we have a special situation known as the natural logarithm function.
Following these rules, we can integrate f'(x) as follows:
∫ f'(x) dx = ∫ ((x^2 - 1) / x^(-1)) dx
= ∫ (x^2 - 1) * x dx
Expanding the expression further:
= ∫ (x^(2+1)) dx - ∫ (1 * x) dx
= ∫ x^3 dx - ∫ x dx
= (x^4 / 4) - (x^2 / 2) + C
Therefore, f(x) = (x^4 / 4) - (x^2 / 2) + C, where C is the constant of integration.