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

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

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

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

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

Decimal

2.01713 \dots

Solution steps

\int_{1}^{4} 5 \cdot \frac{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)

y = 5 sec (5 x)

(5.6) (5.9)

y = 3 x^{2} - x^{3}

(x + y)^{2} d x + (x + y)^{2} d y = 0

y = \frac{10}{x ^{2} - 4 x - 21}, x = - 2, x = 2

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