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

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

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integral from 1 to 4 of (5ln(x))/(x^2)

\int_{1}^{4} \frac{5 ln ( x )}{x ^{2}} d x

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

Solution steps

\int_{1}^{4} \frac{5 ln ( x )}{x ^{2}} d x

Take the constant out:

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

= 5 \cdot \int_{1}^{4} \frac{ln ( x )}{x ^{2}} d x

Apply Integration By Parts

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

= 5 [- \frac{ln ( x )}{x} - \int - \frac{1}{x ^{2}} d x]_{1}^{4}

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

= 5 [- \frac{ln ( x )}{x} - \frac{1}{x}]_{1}^{4}

Since the denominators are equal, combine the fractions:

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

= 5 [\frac{- ln ( x ) - 1}{x}]_{1}^{4}

Compute the boundaries:

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

= 5 (\frac{- 2 ln ( 2 ) - 1}{4} + 1)

\frac{d}{d x} (1 - e^{- x} - x e^{- x})

t y^{^{'}} = y - 5 t^{2} sin (3 t), y (\frac{π}{3}) = 0

\frac{\partial}{\partial x} (e^{2 x} sin (y))

x \to 0 lim (cot (5 x))

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

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