Evaluate $I = \int_{0}^{1} \frac{e^x - 1}{x} dx$ with Taylor Polynomial. 
Evaluate $$I=\int_0^1 \frac{e^x-1}{x} dx.$$

The way I am trying to do this is by substituting $e^x$ with $p_n(x)$ and
$R_n(x)$, the $n$-th Taylor Polynomial and its associated remainder.
Would it be fair to say that 
$$I = \int_0^1 \frac{p_n(x) + R_n(x) - 1}{x}dx $$
and then
the Remainder of $I$ given the current approximation polynomial is
$$ \int_0^1 \frac{R_n(x)}{x}dx?$$
If that's the case then the problem is easy. But if not then I am unsure how to proceed. The hint my book gave me was to expand the exponential with its Taylor polynomial and remainder.
 A: Hint: $e^x=1+x+\frac{x^2}{2}+\frac{x^3}{6}+...=\sum_{n=0}^\infty \frac{x^n}{n!},$ so $e^{x}-1=\sum_{n=1}^\infty\frac{x^n}{n!}$, hence $\frac{e^x-1}{x}=\sum_{n=1}^\infty \frac{x^{n-1}}{{n!}}.$
Then,
$$
\int_{0}^{1}\frac{e^x-1}{x}\,dx=\sum_{n=1}^\infty \int_{0}^{1}\frac{x^{n-1}}{n!}dx=\sum_{n=1}^\infty\bigg[\frac{1}{n}\frac{x^n}{n!}\bigg]_{x=0}^{x=1}=\sum_{n=1}^\infty \frac{1}{n\cdot n!}
$$
A: First note that near $0$ we have : $\frac{e^x-1}{x} \sim 1$ hence your improper integral is well defined. 
Use the fact that : 
$$e^x = \sum_{n = 0}^n \frac{x^n}{n!}$$
Thus : 
$$\frac{e^x - 1}{x} = \sum_{n = 1}^\infty \frac{x^{n-1}}{n!}$$
Hence we get : 
$$\int_0^1 \frac{e^x-1}{x} \mathrm{d}x = \sum_{ n = 1}^\infty \int_0^1 \frac{x^{n-1}}{n!} \mathrm{d}x$$
N.B : it's possible to interchange integrand an sum here since $\forall n \geq 1, \int_0^1 \frac{x^{n-1}}{n!} \mathrm{d}x \leq \frac{1}{n!}$ which is the general term of a convergent serie
A: $$I=\int_0^1 \frac{e^x-1}{x} dx$$
$$=\int_0^1 \frac{\sum_{n=0}^{\infty}\frac{x^n}{n!}-1}{x} dx$$
$$=\int_0^1 \frac{\sum_{n=1}^{\infty}\frac{x^n}{n!}}{x} dx$$
$$=\int_0^1 \sum_{n=1}^{\infty}\frac{x^{n-1}}{n!} dx$$
$$=\sum_{n=1}^{\infty}\int_0^1 \frac{x^{n-1}}{n!} dx$$
$$=\sum_{n=1}^{\infty}\Big[\frac{x^{n}}{n\cdot n!}\Big]_0^1$$
$$=\sum_{n=1}^{\infty}\frac{1}{n\cdot n!}$$
