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What else can I help you with?
What is the indefinite integral of 1 over the Li of x?
Better call it Li2(x).
What is the integral of 1 over x?
ln |x|+C is the answer
What is the integral of 1 over x squared plus 1?
arctan(x)
What is the integral of 3 over 3x?
The 3s would cancel and it would become the integral of 1/x which is ln x.
What you the integral of 1?
if you are integrating with respect to x, the indefinite integral of 1 is just x
Integral of 1 divided by sinx cosx?
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
What is the integral of x divided by x plus 1?
x/(x+1) = 1 - 1/(x + 1), so the antiderivative (or indefinite integral) is x + ln |x + 1| + C,
What is the integral of 1 divided by x squared?
The indefinite integral of (1/x^2)*dx is -1/x+C.
What is the integral of arcsinxdx?
The integral of arcsin(x) dx is x arcsin(x) + (1-x2)1/2 + C.
What is the antiderivative of x to the 1?
By antiderivative do you mean integral? If yes, integral x^1 dx= (x^2)/2
What is the integral of 1 over 1 minus x squared?
∫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)
What is the integral of x diveded by x minus one?
integral x/(x-1) .dx = x - ln(x-1) + c where ln = natural logarithm and c = constant of integration alternatively if you meant: integral x/x - 1 .dx = c
