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

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

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

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

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

Decimal Notation

- 0.76713 \dots

Solution steps

\int_{1}^{2} - \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}^{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}

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

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

= - 5 [- \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}

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

Compute the boundaries:

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

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

\int \frac{- 5}{x} d x

\int x^{- 6} (x + 1) d x

y^{^{''}} - 9 y^{^{'}} + 18 y = 2 sin (5 t)

e^{c o s (s i n (x^{2}))}

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

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