Evaluate the definite integral: int_{1}^{e} ( | vedclass

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(D) Let $I = \int_{1}^{e} (x+1) e^{x} \ln x \, dx$. We know that $\int e^{x} (f(x) + f'(x)) \, dx = e^{x} f(x) + C$. Rewrite the integrand as: $(x+1)

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$e^{e} - 1$

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(D) Let $I = \int_{1}^{e} (x+1) e^{x} \ln x \, dx$. We know that $\int e^{x} (f(x) + f'(x)) \, dx = e^{x} f(x) + C$. Rewrite the integrand as: $(x+1) e^{x} \ln x = e^{x} (x \ln x + \ln x) = e^{x} (x \ln x + \ln x + 1 - 1) = e^{x} ((x \ln x) + (\ln x + 1) - 1)$. This does not immediately fit the form. Let us use integration by parts. Let $u = \ln x$ and $dv = (x+1)e^{x} dx$. Then $du = \frac{1}{x} dx$ and $v = \int (x+1)e^{x} dx = x e^{x}$. Using $\int u \, dv = uv - \int v \, du$: $I = [x e^{x} \ln x]_{1}^{e} - \int_{1}^{e} x e^{x} \cdot \frac{1}{x} \, dx$ $I = [x e^{x} \ln x]_{1}^{e} - \int_{1}^{e} e^{x} \, dx$ $I = [x e^{x} \ln x - e^{x}]_{1}^{e}$ $I = (e \cdot e^{e} \cdot \ln e - e^{e}) - (1 \cdot e^{1} \cdot \ln 1 - e^{1})$ $I = (e \cdot e^{e} \cdot 1 - e^{e}) - (0 - e)$ $I = e^{e+1} - e^{e} + e = e^{e}(e-1) + e$.

Evaluate the definite integral: $\int_{1}^{e} (x+1) e^{x} \ln x \, dx$

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