Apr 5, 2015 · The answer is 0. ln(1) is the same as asking e to what power is 1? Since anything to the 0 power is 1, ln(1) = 0

I found: x=e^e=15.154 You can use the definition of logarithm: log_ax=b->x=a^b and the fact that ln=log_e where e=2.71828...: we can write: ln(ln(x))=1 ln(x)=e^1 x=e^e=15.154

Feb 6, 2017 · ln(1+x) = sum_(n=0)^oo (-1)^n x^(n+1)/(n+1) with radius of convergence R=1 Start from: ln(1+x) = int_0^x (dt)/(1+t) Now the integrand function is the sum of a geometric series of ratio -t: 1/(1+t) = sum_(n=0)^oo (-1)^nt^n so: ln(1+x) = int_0^x sum_(n=0)^oo (-1)^nt^n This series has radius of convergence R=1, so in the interval x in (-1,1) we can integrate term by term: …

Apr 26, 2018 · Then, dv=1 \ dx, v=int1 \ dx=x. We don't put the constant until we finish the whole integration. Inputting, we get, intln(1/x) \ dx=xln(1/x)-intx*(-1/x) \ dx =xln(1/x)-int-1 \ dx =xlnx(1/x)-(-x) =xln(1/x)+x We now simplify the xln(1/x) part. Notice that ln(1/x)=ln(x^-1)=-1lnx=-lnx by the power rule for logarithms.

Nov 14, 2015 · x=(1+sqrt(4e+1))/2 Using the rules of logarithms, ln(x)+ln(x-1)=ln(x*(x-1))=ln(x^2-x). Therefore, ln(x^2-x)=1. Then, we exponentiate both sides (put both sides to the e power): e^(ln(x^2-x))=e^1. Simplify, remembering that exponents undo logarithms: x^2-x=e. Now, we complete the square: x^2-x+1/4=e+1/4 Simplify: (x-1/2)^2 = e+1/4 = (4e+1)/4 Take the square …

Jul 17, 2016 · intln(1+x^2)dx=xln(1+x^2)-2x+arctan(x)+C First, applying integration by parts, we let u = ln(1+x^2) and dv = dx => du = (2x)/(1+x^2) and v = x Applying the formula intudv = uv-intvdu, we have intln(1+x^2)dx = xln(1+x^2)-2intx^2/(1+x^2)dx To solve the remaining integral, we will use trig substitution.

Mar 6, 2007 · Actually, that step is perfectly valid in general - (a/a) ln(x) = 1/a ln(x^a), i.e. when everything is defined. Your next line explains why it's not valid here: If you look at the logarithm graph, you will see that the function is not defined for negative x.

Feb 23, 2006 · the problem reads develop expansion of ln(1+z) of course I just tried throwing it into the formula for taylor expansions, however I do not know what F(a)... Insights Blog -- Browse All Articles -- Physics Articles Physics Tutorials Physics Guides Physics FAQ Math Articles Math Tutorials Math Guides Math FAQ Education Articles Education Guides ...

Jan 14, 2018 · We start by working out a taylor series for #ln(1+x)#. I will be expanding around #0#, so it will be a Maclaurin series. The general formula for a Maclaurin series is: #f(x)=sum_(n=0)^oof^n(0)/(n!)x^n# This means we need to work out the nth derivative of #ln(1+x)#. Let's start by taking some derivatives and see what values they produce at #x=0#:

Aug 29, 2014 · The Maclaurin series of f(x)=ln(1+x) is: f(x)=sum_{n=0}^{infty}(-1)^{n}{x^{n+1}}/{n+1}, where |x|<1. First, let us find the Maclaurin series for f'(x)=1/{1+x}=1/{1 ...