-e-x + C.
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.
-1
One can use integration by parts to solve this. The answer is (x-1)e^x.
If you mean 2x times -2x then it would be -8x.
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.
It is -exp (-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.
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))