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

2015年7月4日 · You can use the definition of logarithm: logax = b → x = ab and the fact that ln = loge where e = 2.71828...: we can write: ln(ln(x)) = 1 ln(x) = e1 x = ee = 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 ...

2018年5月23日 · 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 …

2017年2月6日 · Start from: ln(1 + x) = ∫ x 0 dt 1 +t Now the integrand function is the sum of a geometric series of ratio −t: 1 1 + t = ∞ ∑ n=0(− 1)ntn so: ln(1 + x) = ∫ x 0 ∞ ∑ n=0(−1)ntn This …

2018年3月2日 · 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 …

2016年1月29日 · Note that frac {d} {dx} (ln (1-x)) = frac {-1} {1-x}, x<1. You can express frac {-1} {1-x} as a power series using binomial expansion (for x in the neighborhood of zero). frac {-1} …

2016年6月10日 · We apply the properties of the logarithm: ln(1 e) = ln(e−1) the first property is that the exponent "exit" and multiply the log ln(e−1) = − ln(e) the second property is that the …

2018年4月26日 · To find ∫ln(1/x)dx, we use the integral of inverse functions theorem. Let g be the inverse of a continuous function f. Let F be an antiderivative of f. Then ∫g(x)dx = xg(x) −F (g(x)) …

2017年1月26日 · 1/e lnx=-1=>log_ (e)x=-1 =>e^ (-1)=x :.x=1/e