If the term is -x, the integral expression is simply -∫x. By undoing the power rule, we get -(1/2)x^2+C, an arbitrary constant.
The antiderivative of x2 + x is 1/3x3 + 1/2x2 + C.
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
2x + 1 = 1x + 11Subtract 1x from each side:1x + 1 = 11Subtract 1 from each side:1x = 10
1x1=1
-1
X(logX-1) + C
if is (x-6)-(1x-1) = -5 if id (x-6)-(1x+1) = -7
The antiderivative of 2x is x2.
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.
1
It is ln(ln(x))