Q: What is the antiderivative of exp -x?

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

First, antiderivative = a solution to the indefinite integral therefore to integrate -(csc(x))(cot(x)) first convert it to -cos(x)/sin2(x) To integrate ∫-cos(x)/sin2(x) dx, use substitution u = sin(x) and du/dx = cosx This will make it ∫-1/u2 du and the antiderivative is 1/u +c, therefore the answer is 1/sin(x) + c.

The antiderivative, or indefinite integral, of ex, is ex + C.

The antiderivative of x2 + x is 1/3x3 + 1/2x2 + C.

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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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

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

-e-x + C.

By using the chain rule. Since the derivative of exp(x) is exp(x), the derivative of exp(exp(exp(x))) is exp(exp(exp(x))) times the derivative of what is inside the parentheses, i.e., exp(exp(exp(x))) times derivate of exp(exp(x)). Continue using the chain rule once more, for this expression.

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.

X(logX-1) + C

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.

(2/3)*x^(3/2)

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.

The anti-derivative of any constant c, is just c*x. Thus, the antiderivative of pi is pi*x. We can verify this by taking the derivative of pi*x, which gives us pi.

It is ln(ln(x))