I can't figure this one out, the antiderivative of x^-1.
I thought it would be c, but the derivative of c is zero.
That would be the integral of 1/x. It is called the natural logarithm of x, or ln x.
You can't use the rule that the integral of x^n is x^(n+1)/(n+1) on this one, because n = -1 and you get a zero denominator
Integration of 1/x with respect to x is nayural logarithm of x
Integration of 1/x with respect to x is natural logarithm of x
To find the antiderivative of x^(-1), you can use the power rule for antiderivatives. However, the power rule only applies to functions of the form x^n, where n is a constant. In this case, x^-1 does not fall into that category.
The function x^-1 is better expressed as 1/x, which is also known as the reciprocal function. To find the antiderivative of 1/x, you need to use logarithmic differentiation.
Here's how you can do it:
1. Rewrite the function as 1/x.
2. Recognize that the derivative of ln|x| is 1/x, which closely resembles the function we're trying to find the antiderivative for.
3. Apply the power rule for antiderivatives: ∫1/x dx = ln|x| + C, where C is the constant of integration.
Therefore, the antiderivative of x^-1 (or 1/x) is ln|x| + C.
Please note that in the case of x > 0, the absolute value signs are unnecessary since x is positive. However, when x is negative, the absolute value is required to ensure the function remains continuous and differentiable around zero.