int_{-1}^{0} frac{4x^2 + 4x + 3}{1 + e^{2x + | vedclass

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(C) Let $I = \int_{-1}^{0} \frac{4x^2 + 4x + 3}{1 + e^{2x + 1}} dx$. Notice that $4x^2 + 4x + 3 = (2x + 1)^2 + 2$. Substitute $t = 2x + 1$,then $dt =

Vedclass · Accepted Answer

$\frac{7}{6}$

Vedclass · Answer

(C) Let $I = \int_{-1}^{0} \frac{4x^2 + 4x + 3}{1 + e^{2x + 1}} dx$. Notice that $4x^2 + 4x + 3 = (2x + 1)^2 + 2$. Substitute $t = 2x + 1$,then $dt = 2dx$ or $dx = \frac{dt}{2}$. When $x = -1$,$t = -1$. When $x = 0$,$t = 1$. $I = \int_{-1}^{1} \frac{t^2 + 2}{1 + e^t} \cdot \frac{dt}{2} = \frac{1}{2} \int_{-1}^{1} \frac{t^2 + 2}{1 + e^t} dt$. Using the property $\int_{a}^{b} f(t) dt = \int_{a}^{b} f(a+b-t) dt$,here $a+b = -1+1 = 0$,so $I = \frac{1}{2} \int_{-1}^{1} \frac{(-t)^2 + 2}{1 + e^{-t}} dt = \frac{1}{2} \int_{-1}^{1} \frac{t^2 + 2}{1 + e^{-t}} \cdot \frac{e^t}{e^t} dt = \frac{1}{2} \int_{-1}^{1} \frac{(t^2 + 2)e^t}{e^t + 1} dt$. Adding the two expressions for $I$: $2I = \frac{1}{2} \int_{-1}^{1} \frac{t^2 + 2}{1 + e^t} dt + \frac{1}{2} \int_{-1}^{1} \frac{(t^2 + 2)e^t}{1 + e^t} dt = \frac{1}{2} \int_{-1}^{1} (t^2 + 2) \frac{1 + e^t}{1 + e^t} dt = \frac{1}{2} \int_{-1}^{1} (t^2 + 2) dt$. Since $t^2 + 2$ is an even function,$\int_{-1}^{1} (t^2 + 2) dt = 2 \int_{0}^{1} (t^2 + 2) dt$. $2I = \frac{1}{2} \cdot 2 \int_{0}^{1} (t^2 + 2) dt = \left[ \frac{t^3}{3} + 2t \right]_{0}^{1} = \frac{1}{3} + 2 = \frac{7}{3}$. Thus,$2I = \frac{7}{3} \implies I = \frac{7}{6}$.

$\int_{-1}^{0} \frac{4x^2 + 4x + 3}{1 + e^{2x + 1}} dx = $

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