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ln |x|+C is the answer

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Q: What is the integral of 1 over x?
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Related questions

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 is the integral of 1 over Li of x?

The answer will depend on what "Li of x" means.


What is the indefinite integral of 1 over the Li of x?

Better call it Li2(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 anti-derivative of the absolute value of x over 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.