Best Answer

Unfortunately, whatever symbol that you put between the '2x' and the '1' didn't make it.

Let me surmise that it was either a '+' or a '-'.

+:

x2 + x

-:

x2 - x

If neither of these is what you wanted then let me suggest that you try submitting a question in which you substitute the name of the operator for it symbol; for example, use 'x plus y' instead of 'x+y'.

Q: What is the anti-derivative of 2x 1?

Write your answer...

Submit

Still have questions?

Continue Learning about Math & Arithmetic

I assume you mean -10x^4? In that case, antiderivative would be to add one to the exponent, then divide by the exponent. So -10x^5, then divide by 5. So the antiderivative is -2x^5.

If: x = -3x+1 Then: x+3x = 1 => 4x =1 So: x = 1/4 or 0.25 ----------- I notice that the question requests a solution for g x = -3x + 1. It seems possible that parentheses around the 'x' after the 'g' have gone missing, along with a prime indicating the derivative of the function g. This being the case, we would be seeking the antiderivative of -3x + 1. The antiderivative of a sum is the sum of the antiderivatives. So we can look at -3x and +1 separately. The derivative of x2 is 2x. Therefore, the antiderivative of x is x2/2, and the antiderivative of -3x is -3x2/2. The antiderivative of 1 is x. Overall, the solution is the antiderivative -3x2/2 + x + C, where C is an arbitrary constant.

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

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

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.

Related questions

The antiderivative of 2x is x2.

If you mean 2x times -2x then it would be -8x.

x2-2x+C, where C is some arbitrary constant.

I assume you mean -10x^4? In that case, antiderivative would be to add one to the exponent, then divide by the exponent. So -10x^5, then divide by 5. So the antiderivative is -2x^5.

For example, the derivate of x2 is 2x; then, an antiderivative of 2x is x2. That is to say, you need to find a function whose derivative is the given function. The antiderivative is also known as the indifinite integral. If you can find an antiderivative for a function, it is fairly easy to find the area under the curve of the original function - i.e., the definite integral.

If: x = -3x+1 Then: x+3x = 1 => 4x =1 So: x = 1/4 or 0.25 ----------- I notice that the question requests a solution for g x = -3x + 1. It seems possible that parentheses around the 'x' after the 'g' have gone missing, along with a prime indicating the derivative of the function g. This being the case, we would be seeking the antiderivative of -3x + 1. The antiderivative of a sum is the sum of the antiderivatives. So we can look at -3x and +1 separately. The derivative of x2 is 2x. Therefore, the antiderivative of x is x2/2, and the antiderivative of -3x is -3x2/2. The antiderivative of 1 is x. Overall, the solution is the antiderivative -3x2/2 + x + C, where C is an arbitrary constant.

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

-1

X(logX-1) + C

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).

It is ln(ln(x))