Indefinite Integration 1 Question 28

29. The value of the integral $\int _{e^{-1}}^{e^{2}} \frac{\log _e x}{x} d x$ is

(2000, 2M)

(a) $3 / 2$

(b) $5 / 2$

(c) 3

(d) 5

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Solution:

  1. $\int _{e^{-1}}^{e^{2}} \frac{\log _e x}{x} d x=\int _{e^{-1}}^{1} \frac{\log _e x}{x} d x-\int _1^{e^{2}} \frac{\log _e x}{x} d x$ since, 1 is turning point for $\frac{\log _e x}{x}$ for + ve and - ve values

$$ \begin{aligned} & =-\int _{e^{-1}}^{1} \frac{\log _e x}{x} d x+\int _1^{e^{2}} \frac{\log _e x}{x} d x \\ & =-\frac{1}{2}\left[\left(\log _e x\right)^{2}\right] _{e^{-1}}^{1}+\frac{1}{2}\left[\left(\log _e x\right)^{2}\right] _1^{e^{2}} \\ & =-\frac{1}{2}{0-(-1)^{2} }+\frac{1}{2}\left(2^{2}-0\right)=\frac{5}{2} \end{aligned} $$



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