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

The integral from 1 to e^3 of (ln(x))/(x^2) is -4/(e^3)+1

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integral from 1 to e^3 of (ln(x))/(x^2)

\int_{1}^{e^{3}} \frac{ln ( x )}{x ^{2}} d x

- \frac{4}{e ^{3}} + 1

Decimal

0.80085 \dots

Solution steps

\int_{1}^{e^{3}} \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^{3}}

= [\frac{- ln ( x ) - x \cdot \int - \frac{1}{x ^{2}} d x}{x}]_{1}^{e^{3}}

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

= [\frac{- ln ( x ) - x \frac{1}{x}}{x}]_{1}^{e^{3}}

x \frac{1}{x} = 1

= [\frac{- ln ( x ) - 1}{x}]_{1}^{e^{3}}

Compute the boundaries:

- \frac{4}{e ^{3}} + 1

= - \frac{4}{e ^{3}} + 1

y^{^{'}} + y = e^{t}

\int \frac{x ^{2} - 9}{x ^{3}} d x

g (x) = x

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

\frac{\partial}{\partial x} (2 x^{4} y^{3} + y)

What is the integral from 1 to e^3 of (ln(x))/(x^2) ?
The integral from 1 to e^3 of (ln(x))/(x^2) is -4/(e^3)+1