What is the integral of (ln(x))^6 1/x ?

The integral of (ln(x))^6 1/x is (ln^7(x))/7+C

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integral of (ln(x))^6 1/x

\int (ln (x))^{6} \frac{1}{x} d x

\frac{ln ^{7} ( x )}{7} + C

Solution steps

\int (ln (x))^{6} \frac{1}{x} d x

Simplify

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

= \int \frac{ln ^{6} ( x )}{x} d x

Apply u-substitution

: \int u^{6} d u

= \int u^{6} d u

Apply the Power Rule

: \frac{u ^{7}}{7}

= \frac{u ^{7}}{7}

Substitute back

u = ln (x)

= \frac{ln ^{7} ( x )}{7}

Add a constant to the solution

= \frac{ln ^{7} ( x )}{7} + C

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What is the integral of (ln(x))^6 1/x ?
The integral of (ln(x))^6 1/x is (ln^7(x))/7+C