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

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

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

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

- \frac{4}{e} + 2

Decimal Notation

0.52848 \dots

Solution steps

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

Apply Integration By Parts

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

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

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

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

\frac{- ln ( x ^{2} ) - x \frac{2}{x}}{x} = \frac{- 2 ln ( x ) - 2}{x}

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

Compute the boundaries:

- \frac{4}{e} + 2

= - \frac{4}{e} + 2

\int_{2}^{0} - 4 x^{4} + 8 x^{3} d x

\int_{- 1}^{1} ∣ e^{x} - 1 ∣ d x

\int_{2}^{4} 3 x^{2} d x

\int_{- 1}^{1} x + 2 x^{2} - x^{3} + 5 x^{4} d x

\int_{0}^{1} (x + 3) d x

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