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

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

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

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

- ln (2) + 1

Decimal Notation

0.30685 \dots

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Apply Integration By Parts

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

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

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

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

Since the denominators are equal, combine the fractions:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

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

Compute the boundaries:

\frac{- ln ( 2 ) - 1}{2} + 1

= 2 (\frac{- ln ( 2 ) - 1}{2} + 1)

Simplify

= - ln (2) + 1

\frac{d}{d t} (tan (sec (cos (t))))

\int (1 + x y) d y

n = 0 \sum \infty \frac{1}{3 n + 1}

(\frac{2}{x})^{^{'}}

\int_{- 1}^{1} \frac{1}{2} x^{2} d x

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