The supposed correct answers are: $$\ln(1+x) = \int \left(\frac{1}{1+x}\right)dx$$ $$\ln(1+x) = \sum_{k=0}^{\infty} \int (-x)^k dx$$

In this tutorial we shall derive the series expansion of the trigonometric function $$\ln \left( {1 + x} \right)$$ by using Maclaurin’s series expansion function.

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line\:(1,\:2),\:(3,\:1) f(x)=x^3 ; prove\:\tan^2(x)-\sin^2(x)=\tan^2(x)\sin^2(x) \frac{d}{dx}(\frac{3x+9}{2-x}) (\sin^2(\theta))' \sin(120) \lim _{x\to 0}(x\ln (x)) \int e^x\cos (x)dx

2019年4月19日 · I know you can get $\ln(1-x) \approx -x$ by e.g. substitute $x\rightarrow -x$ into the expansion of $\ln(1+x)$ and through other methods etc. But I still don't quite get how you …

In this guide, we explain the four most important natural logarithm rules, discuss other natural log properties you should know, go over several examples of varying difficulty, and explain how …

2015年7月28日 · I want to prove that: $\ln(x+1)< x$. My idea is to define: $f(x) = \ln(x+1) - x$, so: $f'(x) = \dfrac1{1+x} - 1 = \dfrac{-x}{1+x} < 0, \text{ for }x >0$. Which leads to $f(x)<f(0)$...

Maclaurin series of ln((1+x)/(1-x)).How to expand ln((1+x)/(1-x)) in Maclaurin series?How to expand ln((1+x) over (1-x)) in Maclaurin series?How to obtain th...

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Natural logarithm is the logarithm to the base e of a number. When. Then base e logarithm of x is. The e constant or Euler's number is: The natural logarithm function ln (x) is the inverse …