xln(x)-x
It's either floor(x)+1 or ceil(x) depending on what you want to get for integral x numbers:x or x+1
The limit as x approaches zero of sin(x) over x can be determined using the squeeze theorem.For 0 < x < pi/2, sin(x) < x < tan(x)Divide by sin(x), and you get 1 < x/sin(x) < tan(x)/sin(x)That is the same as 1 < x/sin(x) < 1/cos(x)But the limit as x approaches zero of 1/cos(x) is 1,so 1 < x/sin(x) < 1which means that the limit as x approaches zero of x over sin(x) is 1, and that also means the inverse; the limit as x approaches zero of sin(x) over x is 1.You can also solve this using deriviatives...The deriviative d/dxx is 1, at all points. The deriviative d/dxsin(x) at x=0 is also 1.This means you have the division of two functions, sin(x) and x, at a point where their slope is the same, so the limit reduces to 1 over 1, which is 1.
It depends on the maximum value of c. In signed values, the maximum value we can store in an integral is 2 to the power of the number of bits in the integral, minus 1. Thus a 32-bit signed integral can accommodate all positive values in the range 2^31, which is 2,147,483,648.
The equation for the average over time T is integral 0 to T of I.dt
arctan(x)
The 3s would cancel and it would become the integral of 1/x which is ln x.
The answer will depend on what "Li of x" means.
Better call it Li2(x).
if you are integrating with respect to x, the indefinite integral of 1 is just x
Integral of [1/(sin x cos x) dx] (substitute sin2 x + cos2 x for 1)= Integral of [(sin2 x + cos2 x)/(sin x cos x) dx]= Integral of [sin2 x/(sin x cos x) dx] + Integral of [cos2 x/(sin x cos x) dx]= Integral of (sin x/cos x dx) + Integral of (cos x/sin x dx)= Integral of tan x dx + Integral of cot x dx= ln |sec x| + ln |sin x| + C
x/(x+1) = 1 - 1/(x + 1), so the antiderivative (or indefinite integral) is x + ln |x + 1| + C,
The indefinite integral of (1/x^2)*dx is -1/x+C.
The integral of arcsin(x) dx is x arcsin(x) + (1-x2)1/2 + C.
By antiderivative do you mean integral? If yes, integral x^1 dx= (x^2)/2
∫1/(1-x)2dxWe can rewrite the integral as ∫(1-x)-2dx.Thus:∫(1-x)-2dxu = 1-xdu = -1-1∫-1(1-x)-2dx-1∫u-2du(u-1|0u=1-x(1-x)-1So we conclude that ∫1/(1-x)2dx = (1-x)-1 or 1/(1-x)
For positive x, this expression is equal to 1. The integral (anti-derivative) is therefore x + C (where C is the arbitrary integration constant). For negative x, this expression is equal to -1, and the integral is -x + C. Wolfram Alpha gives the integral as x times sgn(x), where sgn(x) is the "sign" function.