What is the integral from 1 to e^4 of (ln(x))/(x^2) ?

The integral from 1 to e^4 of (ln(x))/(x^2) is -5/(e^4)+1

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integral from 1 to e^4 of (ln(x))/(x^2)

\int_{1}^{e^{4}} \frac{ln ( x )}{x ^{2}} d x

- \frac{5}{e ^{4}} + 1

Decimal Notation

0.90842 \dots

Solution steps

\int_{1}^{e^{4}} \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^{4}}

= [\frac{- ln ( x ) - x \cdot \int - \frac{1}{x ^{2}} d x}{x}]_{1}^{e^{4}}

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

= [\frac{- ln ( x ) - x \frac{1}{x}}{x}]_{1}^{e^{4}}

x \frac{1}{x} = 1

= [\frac{- ln ( x ) - 1}{x}]_{1}^{e^{4}}

Compute the boundaries:

- \frac{5}{e ^{4}} + 1

= - \frac{5}{e ^{4}} + 1

x \to in f ini t o lim (10)

\frac{\partial}{\partial x} (5 (y^{2} - x^{2}) ln (x + y))

t^{2} y^{^{''}} - 2 y = 0

\frac{d}{d x} (sin (x) cos (3 - x^{2}))

\frac{d}{d x} (e^{a x} a)

What is the integral from 1 to e^4 of (ln(x))/(x^2) ?
The integral from 1 to e^4 of (ln(x))/(x^2) is -5/(e^4)+1