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Q: What is the anti-derivative of ln x?
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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.


What is the antiderivative of e to the power of one divided by x?

1/ln(x)*e^(1/x) if you differentiate e^(1/x), you will get ln(x)*e^(1/x). times this by 1/ln(x) and you get you original equation. Peace


Is there any other anti derivative of 1 divided by x?

The antiderivative of 1/x is ln(x) + C. That is, to the natural (base-e) logarithm, you can add any constant, and still have an antiderivative. For example, ln(x) + 5. These are the only antiderivatives; there are no different functions that have the same derivatives. This is valid, in general, for all antiderivatives: if you have one antiderivative of a function, all other antiderivatives are obtained by adding a constant.


What is the integral of x divided by x plus 1?

x/(x+1) = 1 - 1/(x + 1), so the antiderivative (or indefinite integral) is x + ln |x + 1| + C,


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 x divided by x squared minus 1?

The antiderivative of x/(x2-1) is ln(x2-1)/2. Proof: (ln(x2-1)/2)' = (1/(x2-1))*(x2-1)'/2=1/(x2-1)*(2x/2)=x/(x2-1).


What is the antiderivative of x to the 1?

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


What is the antiderivative of x to the negative 6 5ths?

(that weird integral or antiderivative sign) x^(-6/5) dx =-5*x^(-1/5)


What is the antiderivative of e to the -x?

-e-x + C.


What is the antiderivative of exp -x?

It is -exp (-x) + C.


What is the derivative of y equals xlnx?

Use the product rule.y = x lnxy' = x (ln x)' + x' (ln x) = x (1/x) + 1 ln x = 1 + ln xUse the product rule.y = x lnxy' = x (ln x)' + x' (ln x) = x (1/x) + 1 ln x = 1 + ln xUse the product rule.y = x lnxy' = x (ln x)' + x' (ln x) = x (1/x) + 1 ln x = 1 + ln xUse the product rule.y = x lnxy' = x (ln x)' + x' (ln x) = x (1/x) + 1 ln x = 1 + ln x