What is the integral from 1 to 12 of 4/(x^2)ln(x) ?

The integral from 1 to 12 of 4/(x^2)ln(x) is 4((-ln(12)-1)/(12)+1)

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integral from 1 to 12 of 4/(x^2)ln(x)

\int_{1}^{12} \frac{4}{x ^{2}} ln (x) d x

4 (\frac{- ln ( 12 ) - 1}{12} + 1)

Decimal

2.83836 \dots

Solution steps

\int_{1}^{12} \frac{4}{x ^{2}} ln (x) d x

Take the constant out:

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

= 4 \cdot \int_{1}^{12} \frac{1}{x ^{2}} ln (x) d x

Apply Integration By Parts

: [- \frac{ln ( x )}{x} - \int - \frac{1}{x ^{2}} d x]_{1}^{12}

= 4 [- \frac{ln ( x )}{x} - \int - \frac{1}{x ^{2}} d x]_{1}^{12}

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

= 4 [- \frac{ln ( x )}{x} - \frac{1}{x}]_{1}^{12}

Since the denominators are equal, combine the fractions:

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

= 4 [\frac{- ln ( x ) - 1}{x}]_{1}^{12}

Compute the boundaries:

\frac{- ln ( 12 ) - 1}{12} + 1

= 4 (\frac{- ln ( 12 ) - 1}{12} + 1)

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What is the integral from 1 to 12 of 4/(x^2)ln(x) ?
The integral from 1 to 12 of 4/(x^2)ln(x) is 4((-ln(12)-1)/(12)+1)