WBJEE · Maths · Definite Integration
The value of the integral \(\int_{1}^{2} e^{x}\left(\log _{e} x+\frac{x+1}{x}\right) d x\) is
- A \(e^{2}\left(1+\log _{e} 2\right)\)
- B \(e^{2}-e\)
- C \(e^{2}\left(1+\log _{e} 2\right)-e\)
- D \(e^{2}-e\left(1+\log _{e} 2\right)\)
Answer & Solution
Correct Answer
(C) \(e^{2}\left(1+\log _{e} 2\right)-e\)
Step-by-step Solution
Detailed explanation
Let \(I=\int_{1}^{2} e^{x}\left(\log _{e} x+\frac{x+1}{x}\right) d x\) \(\Rightarrow \quad I=\int_{1}^{2}\left(e^{x} \cdot \log _{e} x+e^{x}+\frac{e^{x}}{x}\right) d x\)…
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