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

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

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integral from 1 to 4 of (3ln(x))/(x^2)

\int_{1}^{4} \frac{3 ln ( x )}{x ^{2}} d x

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

Decimal Notation

1.21027 \dots

Solution steps

\int_{1}^{4} \frac{3 ln ( x )}{x ^{2}} d x

Take the constant out:

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

= 3 \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}

= 3 [- \frac{ln ( x )}{x} - \int - \frac{1}{x ^{2}} d x]_{1}^{4}

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

= 3 [- \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}

= 3 [\frac{- ln ( x ) - 1}{x}]_{1}^{4}

Compute the boundaries:

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

= 3 (\frac{- 2 ln ( 2 ) - 1}{4} + 1)

\frac{d}{d x} (\frac{x ^{4}}{9})

y = 2916 x, 108 x^{2}

\int \frac{x ( 2 x - 9 )}{x ^{3} - 6 x ^{2} + 12 x - 8} d x

\int 7 x e^{3 x - 1} d x

\frac{\partial}{\partial x} (cos (y + x))

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