2015年4月5日 · 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

2016年8月20日 · (x+1)ln(1+x)-x+C We have: I=intln(1+x)dx We will use integration by parts, which takes the form: intudv=uv-intvdu So, for intln(1+x)dx, let: {(u=ln(1+x)" "=>" "du=1 ...

2017年2月6日 · 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 …

2018年5月23日 · x=1/(e-1)~~0.582 Step 1 First, we must move all terms to one side. ln(1+x)-1-lnx=0 Step 2 We can now further simplify using the quotient rule. ln((1+x)/x)-1=0 Step 3 We can now combine like terms to reduce the equation. ln(1/x+1)-1=0 Step 4 Next, we begin to isolate the variable, x, by moving everything else to the other side. ln(1/x+1)=1 Step 5 We then use the …

2007年3月6日 · 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.

2018年4月26日 · 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.

2018年3月2日 · -1 Division rule of logarithms states that: ln(x/y) = ln(x) - ln(y) Here we can substitute: ln(1/e)=ln(1) - ln(e) 1) Anything to the power 0=1 2) ln(e)=1, as the base of natural logarithms is always e Here, we can simplify: ln(1)=0 ln(e)=1 Thus: ln(1)-ln(e)=0-1 =-1 Thus, we have our answer

2016年7月17日 · 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.

2016年1月29日 · You can express frac{-1}{1-x} as a power series using binomial expansion (for x in the neighborhood of zero). frac{-1}{1-x} = -(1-x)^{-1} = -( 1 + x + x^2 + x^3 + ... ) To get the Maclaurin Series of ln(1-x), integrate the above "polynomial".