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How do you integrate ln x over square root of x?
x/sqrt(x)=sqrt(x) integral sqrt(x)=2/3x3/2
What is the integral of cot squared x?
ln(sinx) + 1/3ln(sin3x) + C
What is the derivative of x to the power of ln x?
I don't believe that the answer is ln(x)x^(ln(x)-2), since the power rule doesn't apply when you have the variable in the exponent. Do the following instead:y x^ln(x)Taking the natural log of both sides:ln(y)ln(x) * ln(x)ln(y) ln(x)^2Take the derivative of both sides, using the chain rule:1/y * y' 2 ln(x) / xy' 2 ln(x)/ x * yFinally, substitute in the first equation, y x^ln(x):y' 2 ln(x) / x * x^ln(x)y'2 ln(x) * x ^ (ln(x) - 1)Sorry if everything is formatted really badly, this is my first post on answers.com.
How do you integrate cosine squared times sine?
Trying to integrate: cos2x sin x dx Substitute y = cos x Then dy = -sin x dx So the integral becomes: -y2dy Integrating gives -1/3 y3 Substituting back: -1/3 cos3x
What is 2 divided by x as a power of x?
x^(ln(2)/ln(x)-1)
How do you integrate cotxdx?
The integral of cot(x)dx is ln|sin(x)| + C
How do you integrate x power -1?
x-1 = 1/x ∫1/x dx = ln x + C
How do you integrate sinx divide sinx plus cosx?
1/2(x-ln(sin(x)+cos(x)))
What is the derivative of e to the power ln x squared?
e^[ln(x^2)]=x^2, so your question is really, "What is the derivative of x^2," to which the answer is 2x.
How do you integrate 4 over x?
This is the same as 4x-1, so the answer is just 4 ln x + C.
What is the derivative of lnx2?
Do you mean ln(x-2), or ln(x)-2? If it is ln(x-2): 1/(x-2) If it is ln(x)-2: 1/x
How do you integrate 1 over x-1?
2
How do you integrate ln x over square root of x?
x/sqrt(x)=sqrt(x) integral sqrt(x)=2/3x3/2
How do you Differentiate and Integrate y equals 3 to the power of x?
dy/dx = 3^x * ln(3)integral = (3^x) / ln(3)To obtain the above integral...Let y = 3^xln y = x ln 3y = e^(x ln 3)(i.e. 3^x is the same as e^(x ln 3) ).The integral will then be 3^x / ln 3 (from linear composite rule and substitution after integration).
How do you integrate x secx?
To integrate ( x \sec x ), you can use integration by parts. Let ( u = x ) and ( dv = \sec x , dx ). Then, ( du = dx ) and ( v = \ln |\sec x + \tan x| ). Applying the integration by parts formula, you get: [ \int x \sec x , dx = x \ln |\sec x + \tan x| - \int \ln |\sec x + \tan x| , dx + C ] where ( C ) is the constant of integration. The second integral may require further techniques to evaluate.
What is the integral of cot squared x?
ln(sinx) + 1/3ln(sin3x) + C
What is the integral of -2x divided by the square root of x squared minus 4?
- ln ((x^2)-4)
