Solving $\int_0^1 \frac{e^{-t} }{1+t} dt\,$ using Exponential Integral I'm having a really hard time figuring out how to solve the following integrals:
$$
\int_0^1 \frac{e^{-t} }{1+t} dt\, \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 
\int_0^1 \frac{e^{-t} }{(1+t)^2} dt\,
$$
According to my professor, the m-th Exponential Integral is suppose to be used. I'm not that familiar with the Exponential Integral  and am really unsure how to get the above integrals in the right format, particularly for the bounds (Does a transformation need to be done?), or solve it. The equation we were given for the Exponential Integral is $E_m(x)=\int_1^\infty \frac{e^{-xt} }{t^m} dt\,$. 
Any insight would be incredibly helpful (especially steps!!). Thank you!
 A: Note that we can write
$$\begin{align}
\int_0^1 \frac{e^{-t}}{t+1}\,dt&=e\int_1^2 \frac{e^{-t}}{t}\,dt\\\\
&=e\left(\int_1^\infty \frac{e^{-t}}{t}\,dt-\int_2^\infty \frac{e^{-t}}{t}\,dt\right)\\\\
&=e\left(\int_1^\infty \frac{e^{-t}}{t}\,dt-\int_1^\infty \frac{e^{-2t}}{t}\,dt\right)\\\\
&=e\left(E_1(1)-E_1(2)\right)
\end{align}$$
Therefore, we can write

$$\bbox[5px,border:2px solid #C0A000]{\int_0^1 \frac{e^{-t}}{t+1}\,dt=e\left(E_1(1)-E_1(2)\right)}$$


Note that we can write
$$\begin{align}
\int_0^1 \frac{e^{-t}}{(t+1)^2}\,dt&=e\int_1^2 \frac{e^{-t}}{t^2}\,dt\\\\
&=e\left(\int_1^\infty \frac{e^{-t}}{t^2}\,dt-\int_2^\infty \frac{e^{-t}}{t^2}\,dt\right)\\\\
&=e\left(\int_1^\infty \frac{e^{-t}}{t}\,dt-\frac12 \int_1^\infty \frac{e^{-2t}}{t}\,dt\right)\\\\
&=e\left(E_2(1)-\frac12 E_2(2)\right)
\end{align}$$
Therefore, we can write

$$\bbox[5px,border:2px solid #C0A000]{\int_0^1 \frac{e^{-t}}{(t+1)^2}\,dt=e\left(E_1(1)-E_1(2)\right)}$$

A: How about using the geometric series? That is
$$\frac{1}{1+t}=\sum_{n=0}^\infty (-t)^n$$
Then we have 
$$
\int^1_0\frac{e^{-t}}{1+t}\,dt = \sum_{n=0}^\infty \int^1_0 e^{-t}(-t)^n\,dt.
$$
This isn't quite the exponential integral, since the power of $t$ is in the numerator instead of the denominator, and the bounds are $0\leq t\leq1$ instead of $1\leq t\leq\infty$. So let's invert $t$ to transform that. Let $u=\frac{1}{t}$ and we have:
$$
\int^1_0e^{-t}t^n\,dt = -\int^1_\infty \frac{e^{-1/u}}{u^n}\frac{1}{u^2}\,du = \int^\infty_1 \frac{e^{-1/u}}{u^{n+2}}\,du
$$
