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

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

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

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

\frac{2}{x}

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Apply u-substitution

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

= \frac{1}{x} \cdot \int_{0}^{2} u d u

Apply the Power Rule

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

= \frac{1}{x} [\frac{u ^{2}}{2}]_{0}^{2}

Simplify

\frac{1}{x} [\frac{u ^{2}}{2}]_{0}^{2} : \frac{[ \frac{u ^{2}}{2} ] _{0}^{2}}{x}

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

Compute the boundaries:

2

= \frac{2}{x}

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