Answer
$\approx -0.3633060$ (with an error of about $10^{-5}$)
Work Step by Step
We need to integrate the integral with respect to $ x $ as follows:
$\int_0^{0.4} \dfrac{e^{-x}-1}{x} dx=\int_0^{0.4} [-1+\dfrac{x}{2!}-\dfrac{x^{2}}{3!}+\dfrac{x^4}{4!}-...] dx $
Re-arrange as: $\int_0^{0.4} [-1+\dfrac{x}{2!}-\dfrac{x^{2}}{3!}+\dfrac{x^4}{4!}-...] dx=-[x]_0^{0.4}+[\dfrac{x^2}{2 \cdot 2!}]_0^{0.4}-[\dfrac{x^3}{3 \cdot 3!}]_0^{0.4}+......$
$=-0.4+\dfrac{(0.4)^2}{2 \cdot 2!}-\dfrac{(0.4)^3}{3 \cdot 3!}+\dfrac{(0.4)^4}{4 \cdot 4!}-\dfrac{(0.4)^5}{5 \cdot 5!}+\dfrac{(0.4)^6}{6 \cdot 6!}....$
and $\dfrac{(0.4)^6}{6 \cdot 6!} \approx 9.48148 \times 10^{-7} $ so, the proceeding terms will be greater than $10^{-5}$.
This means that the first five terms of the series would give the accuracy.
Therefore, $\int_0^{0.4} \dfrac{e^{-x}-1}{x} dx \approx -0.3633060$ (with an error of about $10^{-5}$)
