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Explain Solution R.D.Sharma Class 12 Chapter 19 Definite Integrals  Exercise 19.1 Question 34 Maths Textbook Solution.

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Hint: Use indefinite formula and put limits to solve this integral


\int_{1}^{e} \frac{e^{x}}{x}(1+x \log x) d x


\begin{aligned} &\int_{1}^{e} \frac{e^{x}}{x}(1+x \log x) d x=\int_{1}^{e}\left(\frac{e^{x}}{x}+\frac{e^{x}}{x} \log x\right) d x \\ &=\int_{1}^{e} \frac{e^{x}}{x} d x+\int_{1}^{e} e^{x} \log x d x \end{aligned}

applying integration by parts method in 1st integral then

\begin{aligned} &=\left[e^{x} \int \frac{1}{x} d x\right]_{1}^{e}-\int_{1}^{e}\left[\frac{d}{d x}\left(e^{x}\right) \int \frac{1}{x} d x\right] d x+\int_{1}^{e} e^{x} \log x d x \\ &=\left[e^{x} \log x\right]_{1}^{e}-\int_{1}^{e} e^{x} \log x d x+\int_{1}^{e} e^{x} \log x d x \\ &=\left[e^{e} \log e-e^{1} \log 1\right] \\ &{[\because \log e=1,1 \log 1=0]} \\ &=\left[e^{e} .1-0\right] \end{aligned}


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