What is the integral from 1 to e^5 of (ln(x))/(x^2) ?

The integral from 1 to e^5 of (ln(x))/(x^2) is -6/(e^5)+1

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integral from 1 to e^5 of (ln(x))/(x^2)

\int_{1}^{e^{5}} \frac{ln ( x )}{x ^{2}} d x

- \frac{6}{e ^{5}} + 1

Decimal

0.95957 \dots

Solution steps

\int_{1}^{e^{5}} \frac{ln ( x )}{x ^{2}} d x

Apply Integration By Parts

= [\frac{- ln ( x ) - x \cdot \int - \frac{1}{x ^{2}} d x}{x}]_{1}^{e^{5}}

\int - \frac{1}{x ^{2}} d x = \frac{1}{x}

= [\frac{- ln ( x ) - x \frac{1}{x}}{x}]_{1}^{e^{5}}

x \frac{1}{x} = 1

= [\frac{- ln ( x ) - 1}{x}]_{1}^{e^{5}}

Compute the boundaries:

- \frac{6}{e ^{5}} + 1

= - \frac{6}{e ^{5}} + 1

What is the integral from 1 to e^5 of (ln(x))/(x^2) ?
The integral from 1 to e^5 of (ln(x))/(x^2) is -6/(e^5)+1