Answer
$\approx -0.3633060$ with an error of about $10^{-5}$
Work Step by Step
We integrate the integral with respect to $ x $ taken a lower limit as $0$ and upper limit as $0.4$.
$\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!}+\dfrac{x^5}{5!}...] dx $
or, $=-[x]_0^{0.4}+[\dfrac{x^2}{2 \cdot 2!}]_0^{0.4}-[\dfrac{x^3}{3 \cdot 3!}]_0^{0.4}+......$
or, $=-0.4+\dfrac{(0.4)^2}{2 \cdot 2!}-\dfrac{(0.4)^3}{3 \cdot 3!}+\dfrac{(0.4)^4}{4 \cdot 4!}-\dfrac{(0.4)^5}{5 \cdot 5!}+\dfrac{(0.4)^6}{6 \cdot 6!}....$
But $\dfrac{(0.4)^6}{6 \cdot 6!} \approx 9.48148 \times 10^{-7} $ and the proceeding term is greater than $10^{-5}$.
so, the first five terms of the series would give the accuracy.
$\int_0^{0.4} \dfrac{e^{-x}-1}{x} dx \approx -0.3633060$ with an error of about $10^{-5}$.
