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For positive x, this expression is equal to 1. The integral (anti-derivative) is therefore x + C (where C is the arbitrary integration constant).

For negative x, this expression is equal to -1, and the integral is -x + C.

Wolfram Alpha gives the integral as x times sgn(x), where sgn(x) is the "sign" function.

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Q: What is the anti-derivative of the absolute value of x over x?

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It is ln(ln(x))

Absolute value of -x is x.

The general formula for powers doesn't work in this case, because there will be a zero in the denominator. The antiderivative of 1/x is ln(x), that is, the natural logarithm of x.

The absolute value of 19 is 19. If x is positive , absolute x equals x.

absolute value

You can't, unless it's an initial value problem. If f(x) is an antiderivative to g(x), then so is f(x) + c, for any c at all.

zero. The absolute value of a number is just the positive version of that number, so the absolute value of x is x, and x minus x is zero.

Antiderivative of x/-1 = -1(x^2)/2 + C = (-1/2)(x^2) + C Wolfram says antiderivative of x^-1 is log(x) + C

By antiderivative do you mean integral? If yes, integral x^1 dx= (x^2)/2

The absolute value of a number is its distance, on the number line, of the number from 0. If x is non-negative, its absolute value is x; and if x is negative then the absolute value is -x (which is positive).

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The absolute value of a number is how many spaces the number is away from 0. So if the number was 32, the absolute value would be 32. And if the number was -54, then the absolute value would be 54. ========== The definition of "absolute value" for a number x (written as |x| ) is: |x| = x for x >0 |x| = 0 for x=0 |x| = -x for x<0

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