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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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Related questions

What is the antiderivative of 1 over xlnx?

It is ln(ln(x))


How do you take the antiderivative of 1 over 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.


How can you calculate the arbitrary constant in the solution to an antiderivative?

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.


What is the absolute value of -19?

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


What does the symbol x with an arrow over it stand for?

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What is the absolute value of x minus the absolute value of x?

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.


What is the antiderivative of x to the -1?

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


What is the anti derivative of 1 divided by x?

If f(x)=1/x then F(x)=antiderivative of f(x)=ln(|x|) (the natural log of the absolute value of x) There's another way of reading this question. The anti derivative of 1 is x+c. Dividing that by x gives you 1 + c/x


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What is the domain of f of x equals x over the absolute value of x - 3 then - 1?

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What is the absolute value of a negative integers?

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