Evaluate given integrals in terms of the integral from 0 to 1 of et / (t+1)

Define an integral function

$A = \int_0^1 \frac{e^t}{t-a-1} \, dt.$

In terms of $A$ evaluate the following:

First, we make the substitution $s = -t+a$ so $ds = -dt$ . The bounds of integration are then
$-(a-1) + a = 1 \qquad \text{and} \qquad -a + a = 0.$

Therefore we have,

$\begin{align*} \int_{a-1}^a \frac{e^{-t}}{t-a-1} \, dt &= -\int_1^0 \frac{e^{s-a}}{-s-1} \, ds \\ &= e^{-a} \int_1^0 \frac{e^s}{s+1} \, ds \\ &= -e^{-a} \int_0^1 \frac{e^s}{s+1} \, ds \\ &= -Ae^{-a}. \end{align*}$
For this one, make the substitution $s = t^2$ , $ds = 2t \, dt$ . The bounds of integration don’t change since $0^2 = 0$ and $1^2 = 1$ . So we have,
$\begin{align*} \int_0^1 \frac{te^{t^2}}{t^2+1} \, dt &= \frac{1}{2} \int_0^1 \frac{2te^{t^2}}{t^2+1} \, dt \\ &= \frac{1}{2} \int_0^1 \frac{e^s}{s+1} \, ds \\ &= \frac{1}{2} A. \end{align*}$
To compute this in terms of $A$ , we integrate by parts. Let
$\begin{align*} u &= e^t & du &= e^t dt \\ dv &= \frac{1}{(t+1)^2} \, dt & v &= \frac{-1}{t+1}. \end{align*}$

Therefore we have

$\begin{align*} \int_0^1 \frac{e^t}{(t+1)^2} \, dt &= -\frac{e^t}{t+1} \Bigr \rvert_0^1 + \int_0^1 \frac{e^t}{t+1} \, dt \\ &= -\frac{e}{2} + 1 + A \\ &= A + 1 - \frac{1}{2}e. \end{align*}$
We use integration by parts again, this time let
$\begin{align*} u &= \log(1+t) & du &= \frac{1}{t+1} \, dt \\ dv &= e^t \, dt & v &= e^t. \end{align*}$

Therefore we have,

$\begin{align*} \int_0^1 e^t \log(1+t) \, dt &= e^t \log(1+t) \Bigr \rvert_0^1 - \int_0^1 \frac{e^t}{t+1} \, dt \\ &= e \log 2 - A. \end{align*}$