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

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If f' (x) = x43, then f(x) = (1/44) x44 + C.

One can use integration by parts to solve this. The answer is (x-1)e^x.

-e-x + C.

The integral would be 10e(1/10)x+c

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It is ln(ln(x))

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

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.

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)

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.

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Powers of e are simple to integrate. The derivative of eu equals u'eu; inversely, the antiderivative of eu equals eu/u'. Therefore, the antiderivative of e1/-x equals (e1/-x)/{d/dx[1/-x]}. The derivative of 1/-x, which can also be expressed as x-1, equals (-1)x(-1-1) = -x-2 = -1/x2.

X(logX-1) + C

x(pi+1)/(pi+1)

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.

sqrt(1) + 3*sqrt(x) = 1 + 3*x^1/2So the antiderivative is x + [3*x^(3/2)]/(3/2) + c = x + 2*x^(3/2) + c where c is the constant of integration.