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

The integral from 1 to e^2 of (ln(x))/(x^2) is -3/(e^2)+1

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

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

- \frac{3}{e ^{2}} + 1

Solution steps

\int_{1}^{e^{2}} \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^{2}}

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

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

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

x \frac{1}{x} = 1

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

Compute the boundaries:

- \frac{3}{e ^{2}} + 1

= - \frac{3}{e ^{2}} + 1

\frac{\partial}{\partial x} (\frac{a - b ^{2}}{x - c ^{2}})

\frac{d}{d x} (5 - \frac{4}{x})

\frac{- u ^{2} + 7}{( u ^{2} + 7 ) ^{2}}

\frac{d}{d x} (sec (x^{5} + x))

(1.1) (0.5)

What is the integral from 1 to e^2 of (ln(x))/(x^2) ?
The integral from 1 to e^2 of (ln(x))/(x^2) is -3/(e^2)+1