What is the integral from 6 to 36 of 1/(x(ln(x))^5) ?

The integral from 6 to 36 of 1/(x(ln(x))^5) is (15)/(64ln^4(6))

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integral from 6 to 36 of 1/(x(ln(x))^5)

\int_{6}^{36} \frac{1}{x ( ln ( x ) ) ^{5}} d x

\frac{15}{64 ln ^{4} ( 6 )}

Decimal Notation

0.02274 \dots

Solution steps

\int_{6}^{36} \frac{1}{x ( ln ( x ) ) ^{5}} d x

Apply u-substitution

: \int_{l n (6)}^{2 l n (6)} \frac{1}{u ^{5}} d u

= \int_{l n (6)}^{2 l n (6)} \frac{1}{u ^{5}} d u

Apply the Power Rule

: [- \frac{1}{4 u ^{4}}]_{l n (6)}^{2 l n (6)}

= [- \frac{1}{4 u ^{4}}]_{l n (6)}^{2 l n (6)}

Compute the boundaries:

\frac{15}{64 ln ^{4} ( 6 )}

= \frac{15}{64 ln ^{4} ( 6 )}

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What is the integral from 6 to 36 of 1/(x(ln(x))^5) ?
The integral from 6 to 36 of 1/(x(ln(x))^5) is (15)/(64ln^4(6))