What is the integral from 1 to e^2 of (ln(x))/x ?

The integral from 1 to e^2 of (ln(x))/x is 2

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integral from 1 to e^2 of (ln(x))/x

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

2

Solution steps

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

Apply u-substitution

= \int_{0}^{2} u d u

Apply the Power Rule

= [\frac{u ^{2}}{2}]_{0}^{2}

Compute the boundaries:

2

= 2

\int_{0}^{\frac{π}{4}} x sin (x) d x

\int_{0}^{2 π} x cos (x) d x

\int_{- 2}^{1} (3 + y^{2}) - (2 - y^{2}) d y

\int_{- 2}^{1} ∣ x + 1 ∣ d x

\int_{0}^{π} (sin (x))^{3} d x

What is the integral from 1 to e^2 of (ln(x))/x ?
The integral from 1 to e^2 of (ln(x))/x is 2