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Better call it Li2(x).
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∙ 11y ago
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What you the integral of 1?
if you are integrating with respect to x, the indefinite integral of 1 is just 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 divided by 2x squared?
0.5
What is the integral of tan squared x cosine to the 5th power x dx?
Since you did not specify any limits of integration, I assume you are looking for the indefinite integral of this expression: tan2(x)cos5(x) with respect to x (dx). Using the following identity: tan(x) = sin(x) / cos(x) The original expression can be rewritten as: (sin2(x) / cos2(x))cos5(x) Which further simplifies to: sin2(x)cos3(x) Which can be expanded to: sin2(x)cos2(x)cos(x) Using the identity: sin2(x) + cos2(x) = 1 which implies: cos2(x) = 1 - sin2(x) which makes the expression from above able to be simplified into: sin2(x)(1 - sin2(x))cos(x) From here, you can use u-substitution by using the substitution: u = sin(x) du = cos(x) dx => dx = du/cos(x) So after u substitution: int(sin2(x)(1 - sin2(x))cos(x)) dx becomes: int(u2(1-u2)) du int(u2-u4) du From here, elementary antiderivatives can be used: anti(u2) = (1/3)(u3) anti(u4) = (1/5)(u5) which yields a final indefinite integral in u of: (1/3)u3-(1/5)u5 + C where C is the constant of integration (since this is an indefinite integral). Back-substituting with the u-substitution from before (u=sin(x)), the final indefinite integral in x is: (1/3)sin3(x) - (1/5)sin5(x) + 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 Li of x?
The answer will depend on what "Li of x" means.
What you the integral of 1?
if you are integrating with respect to x, the indefinite integral of 1 is just x
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 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 definite integral of 1 divided by x squared?
For it to be a definite integral, you would need to specify a range. We can however give you the indefinite integral. The easiest way to do this is to think of it not as a fraction, but as a negative exponent: 1/x2 = x-2 It then becomes quite easy to integrate, as we can say in general: ∫(axn) dx = ax(n + 1) / (n + 1) + C In this case then, we have: ∫(x-2) dx = -x-1 + C, or -1/x + C
How do you evaluate definite integrals?
In order to evaluate a definite integral first find the indefinite integral. Then subtract the integral evaluated at the bottom number (usually the left endpoint) from the integral evaluated at the top number (usually the right endpoint). For example, if I wanted the integral of x from 1 to 2 (written with 1 on the bottom and 2 on the top) I would first evaluate the integral: the integral of x is (x^2)/2 Then I would subtract the integral evaluated at 1 from the integral evaluated at 2: (2^2)/2-(1^2)/2 = 2-1/2 =3/2.
What is the indefinite integral of 2 over X?
S 2/x d/x bring the constant 2 out in front of the sign of integration 2 S 1/x dx you should know the integration of 1/x 2*ln(x) + C
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 is the integral of cosx divided by sinx plus cosx from 0 to 2pi?
The Integral diverges. It has singularities whenever sin(x)+cos(x)=0. Singularities do not necessarily imply that the integral goes to infinity, but that is the case here, since the indefinite integral is x/2 + 1/2 Log[-Cos[x] - Sin[x]]. Obviously this diverges when evaluated at zero and 2pi.
What is the integral of 1 divided by 2x squared?
0.5
