int(ln(x2)dx)=xln|x2|-2x int(ln2(x)dx)=x[(ln|x|-2)ln|x|+2]
You can also write this as ln(6 times 4)
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.
The derivative of e^u(x) with respect to x: [du/dx]*[e^u(x)]For a general exponential: b^x, can be rewritten as b^x = e^(x*ln(b))So derivative of b^x = derivative of e^u(x), where u(x) = x*ln(b).Derivative of x*ln(b) = ln(b). {remember b is just a constant, so ln(b) is a constant}So derivative of b^x = ln(b)*e^(x*ln(b))= ln(b) * b^x(from above)
x^(ln(2)/ln(x)-1)
The integral of cot (x) dx is ln (absolute value (sin (x))) + C. Without using the absolute value, you can use the square root of the square, i.e. ln (square root (sin2x)) + C
This is the same as 4x-1, so the answer is just 4 ln x + C.
2
ln(x times root(y) / z5)ln(x times root(y) / z5)ln(x times root(y) / z5)ln(x times root(y) / z5)
int(ln(x2)dx)=xln|x2|-2x int(ln2(x)dx)=x[(ln|x|-2)ln|x|+2]
- ln ((x^2)-4)
if x8 = 20 then x = - eighth root of 20 or x = eighth root of 20 eighth root of 20 = 1.4542154.... Maybe you mean 8x = 20 If so Ln (8x) = Ln (20) => x Ln(8)= Ln(20) => x = Ln(20)/Ln(8) = 1.4406426...
The integral of cot(x)dx is ln|sin(x)| + C
Pi^2 times the square root of 0.3 = Ln Xi for Ln Yi it's a little tougher. (0.3^4 times the value of Ln Xi)/17.2 Hope this helped. (not)
sqrt(32) = 4sqrt(2) ln surd form Numerically it is 5.656854249....~ 5.66
The integral for csc(u)dx is -ln|csc(u) + cot(u)| + C.
x-1 = 1/x ∫1/x dx = ln x + C