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

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

Answers (1)

Answer:

            e^{e}

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

Given:

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

Solution:

\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}

=e^{e}

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