Note that $$\frac{1}{1+x}=\sum_{n \ge 0} (-1)^nx^n$$ Integrating both sides gives you \begin{align} \ln(1+x) &=\sum_{n \ge 0}\frac{(-1)^nx^{n+1}}{n+1}\\ &=x-\frac{x^2 ...

2019年4月19日 · Stack Exchange Network. Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for …

Stack Exchange Network. Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their …

2015年7月28日 · Stack Exchange Network. Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for …

2017年5月20日 · Evaluate $$\int_{0}^{1} \ln (x) \ln(1-x) dx$$ $\begingroup$ Welcome to math.SE: since you are new, I wanted to let you know a few things about the site.

There are at least two possible methods, the first is by studying the functions' variation (see Timbuc's answer above) and the second is by Integration: $$\forall x>0,\qquad \frac 1x \leq 1 …

2018年5月7日 · Using this power series, $$\frac { 1 }{ 1-x } =\sum _{ k=0 }^{ \infty }{ x^ k }$$ a) find a power series for $\ln(1 − x)$.

2018年10月27日 · This can be solved either by using Lambert W function or Newton Raphson method.The above equation can be written as -> 1 = x*ln(x)

2014年9月18日 · Evaluate the integral, $$ \int_{0}^{1} \ln(x)\ln(1-x)\,dx$$ I solved this problem, by writing power series and then calculating the series and found the answer to be $ 2 -\zeta(2) $, …

Approximating $\ln(1+2x) = 2x$ using linear approximation Hot Network Questions We have Starlink, our location shows incorrectly as Chicago but we live in Missouri.