Antiderivative of x/-1 = -1(x^2)/2 + C = (-1/2)(x^2) + C
Wolfram says antiderivative of x^-1 is log(x) + C
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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.
It is -exp (-x) + 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
The antiderivative, or indefinite integral, of ex, is ex + C.