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Question

Mathematics Question on integral

∫1elog⁡ x dx=\int\limits_1^e \log \, x \, dx =

A

11

B

e−1e - 1

C

e+1e + 1

D

00

Answer

11

Explanation

Solution

We have, ∫1elog⁡ x dx=\int\limits_1^e \log \, x \, dx =
=[log⁡x∫1dx]1e−∫1e(ddxlog⁡ x ∫1 dx)dx=\left[\log x \int\limits 1 dx \right]^{e}_{1} - \int\limits_1^e \left( \frac{d}{dx} \log \, x \, \int\limits 1 \, dx \right) dx
=[log⁡ x.x]1e−∫1e1xx dx=\left[\log\, x . x \right]^{e}_{1} - \int\limits_1^e \frac{1}{x} x \, dx
=[e log⁡ e−1 log⁡(1)]−∫1e1 dx= [e \, \log \, e - 1 \, \log (1) ] - \int\limits_1^e 1 \, dx
=e−[x]1e=e−[e−1]=1= e - [x]^e_1 =e - [e - 1] = 1