Use $\frac{1}{x+1}$ to show that $\ln(x+1) = x - \frac{x^2}{2} + \frac{x^3}{3} - \dots + (-1)^{n-1} \frac{x^n}{n} + \int_{0}^{x} \frac{(-t)^n}{1+t}$

Question

If $x \geq 0$ and $n \in \mathbb{N}$, show that \begin{align*} \frac{1}{x+1} = 1 - x + x^2 - x^3 + \dots + (-x)^{n-1} + \frac{(-x)^n}{1+x} \end{align*} Use this to show that $$\ln(x+1) = x - \frac{x^2}{2} + \frac{x^3}{3} - \dots + (-1)^{n-1} \frac{x^n}{n} + \int_{0}^{x} \frac{(-t)^n}{1+t}dt$$

I am not sure how to get started. This section is about exponentials and I vaguely see taylor's expansino being used in the first equality, but not sure where the last term $\frac{(-x)^n}{1+x}$ comes from. I also see that there might be an inverse relationship at play because there is a logarithm. I am trying to relate this to a problem I might have encountered in calculus, but I can't think of anything either.

Answer 1

hint

You need to know that for $a\ne 1,$

$$1+a+a^2+a^3+...+a^{n-1} = \frac{1-a^n}{1-a}$$

apply with $ a=-x$