What is the integral from 1 to 2 of (10ln(x))/(x^2) ?

The integral from 1 to 2 of (10ln(x))/(x^2) is -5ln(2)+5

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integral from 1 to 2 of (10ln(x))/(x^2)

\int_{1}^{2} \frac{10 ln ( x )}{x ^{2}} d x

- 5 ln (2) + 5

Decimal

1.53426 \dots

Solution steps

\int_{1}^{2} \frac{10 ln ( x )}{x ^{2}} d x

Take the constant out:

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

= 10 \cdot \int_{1}^{2} \frac{ln ( x )}{x ^{2}} d x

Apply Integration By Parts

= 10 [- \frac{ln ( x )}{x} - \int - \frac{1}{x ^{2}} d x]_{1}^{2}

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

= 10 [- \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}

= 10 [\frac{- ln ( x ) - 1}{x}]_{1}^{2}

Compute the boundaries:

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

= 10 (\frac{- ln ( 2 ) - 1}{2} + 1)

Simplify

= - 5 ln (2) + 5

f (x) = arctan (5 x)

\int \frac{x}{x - 11} d x

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What is the integral from 1 to 2 of (10ln(x))/(x^2) ?
The integral from 1 to 2 of (10ln(x))/(x^2) is -5ln(2)+5