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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:

  1. \displaystyle{ \int_{a-1}^a \frac{e^{-t}}{t-a-1} \, dt}.
  2. \displaystyle{ \int_0^1 \frac{te^{t^2}}{t^2+1} \, dt}.
  3. \displaystyle{ \int_0^1 \frac{e^t}{(t+1)^2} \, dt}.
  4. \displaystyle{ \int_0^1 e^t \log (1+t) \, dt}.

  1. 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*}

  2. 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*}

  3. 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*}

  4. 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*}

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