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

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

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

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

- 2 ln (2) + 2

Decimal

0.61370 \dots

Solution steps

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

Take the constant out:

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

= 4 \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}

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

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

= 4 [- \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}

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

Compute the boundaries:

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

= 4 (\frac{- ln ( 2 ) - 1}{2} + 1)

Simplify

= - 2 ln (2) + 2

\frac{d}{d x} (\frac{e ^{x} - e ^{- x}}{2})

\frac{\partial}{\partial x} (cos (y^{3} - 6 x))

e^{- \frac{x}{2}}

\frac{d}{d x} (\frac{x + 3}{x + 1})

\int 7 - 4 x - 4 x^{2} d x

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