What is the integral of ((ln(u))^8)/u ?

The integral of ((ln(u))^8)/u is (ln^9(u))/9+C

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integral of ((ln(u))^8)/u

\int \frac{( ln ( u ) ) ^{8}}{u} d u

\frac{ln ^{9} ( u )}{9} + C

Solution steps

\int \frac{( ln ( u ) ) ^{8}}{u} d u

Apply u-substitution

: \int v^{8} d v

= \int v^{8} d v

Apply the Power Rule

: \frac{v ^{9}}{9}

= \frac{v ^{9}}{9}

Substitute back

v = ln (u)

= \frac{ln ^{9} ( u )}{9}

Add a constant to the solution

= \frac{ln ^{9} ( u )}{9} + C

\frac{d y}{d x} = \frac{e ^{3 x} + 3 x y + y}{x}

\frac{d ^{2}}{d x ^{2}} (\frac{3 x - 2}{2 x + 1})

\frac{d}{d x} (\frac{lo g _{3} ( x )}{lo g _{7} ( x )})

\int x^{2} - cos (x) d x

d x + e^{5 x} d y = 0

What is the integral of ((ln(u))^8)/u ?
The integral of ((ln(u))^8)/u is (ln^9(u))/9+C