What is the integral from 1 to 4 of 1/(x^2)ln(6x) ?

The integral from 1 to 4 of 1/(x^2)ln(6x) is (-ln(24)-1)/4+ln(6)+1

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integral from 1 to 4 of 1/(x^2)ln(6x)

\int_{1}^{4} \frac{1}{x ^{2}} ln (6 x) d x

\frac{- ln ( 24 ) - 1}{4} + ln (6) + 1

Decimal Notation

1.74724 \dots

Solution steps

\int_{1}^{4} \frac{1}{x ^{2}} ln (6 x) d x

Simplify

\frac{1}{x ^{2}} ln (6 x) : \frac{ln ( 6 x )}{x ^{2}}

= \int_{1}^{4} \frac{ln ( 6 x )}{x ^{2}} d x

Apply Integration By Parts

: [- \frac{ln ( 6 x )}{x} - \int - \frac{1}{x ^{2}} d x]_{1}^{4}

= [- \frac{ln ( 6 x )}{x} - \int - \frac{1}{x ^{2}} d x]_{1}^{4}

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

= [- \frac{ln ( 6 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}

= [\frac{- ln ( 6 x ) - 1}{x}]_{1}^{4}

Compute the boundaries:

\frac{- ln ( 24 ) - 1}{4} + ln (6) + 1

= \frac{- ln ( 24 ) - 1}{4} + ln (6) + 1

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What is the integral from 1 to 4 of 1/(x^2)ln(6x) ?
The integral from 1 to 4 of 1/(x^2)ln(6x) is (-ln(24)-1)/4+ln(6)+1