What is the integral from 1 to 2 of log_{e}(x) ?

The integral from 1 to 2 of log_{e}(x) is 2ln(2)-1

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integral from 1 to 2 of log_{e}(x)

\int_{1}^{2} lo g_{e} (x) d x

2 ln (2) - 1

Decimal Notation

0.38629 \dots

Solution steps

\int_{1}^{2} lo g_{e} (x) d x

Apply Integration By Parts

: [x ln (x) - \int 1 d x]_{1}^{2}

= [x ln (x) - \int 1 d x]_{1}^{2}

\int 1 d x = x

= [x ln (x) - x]_{1}^{2}

Compute the boundaries:

2 ln (2) - 1

= 2 ln (2) - 1

\int_{0}^{1} x^{2} x - 1 d x

\int_{0}^{\frac{π}{2}} (e^{t} + cos (t)) d t

\int_{0}^{1} x 2 - x d x

\int_{0}^{\frac{1}{2}} (2 x + 1) e^{x} d x

\int_{3}^{4} \frac{1}{25 - x ^{2}} d x

What is the integral from 1 to 2 of log_{e}(x) ?
The integral from 1 to 2 of log_{e}(x) is 2ln(2)-1