What is the integral of ((ln(x^9)))/x ?

The integral of ((ln(x^9)))/x is 9/2 ln^2(x)+C

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integral of ((ln(x^9)))/x

\int \frac{( ln ( x ^{9} ) )}{x} d x

\frac{9}{2} ln^{2} (x) + C

Solution steps

\int \frac{ln ( x ^{9} )}{x} d x

Apply u-substitution

: \int \frac{ln ( u )}{9 u} d u

= \int \frac{ln ( u )}{9 u} d u

Take the constant out:

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

= \frac{1}{9} \cdot \int \frac{ln ( u )}{u} d u

Apply u-substitution

: \int v d v

= \frac{1}{9} \cdot \int v d v

Apply the Power Rule

: \frac{v ^{2}}{2}

= \frac{1}{9} \cdot \frac{v ^{2}}{2}

Substitute back

= \frac{1}{9} \cdot \frac{ln ^{2} ( x ^{9} )}{2}

Simplify

\frac{1}{9} \cdot \frac{ln ^{2} ( x ^{9} )}{2} : \frac{9}{2} ln^{2} (x)

= \frac{9}{2} ln^{2} (x)

Add a constant to the solution

= \frac{9}{2} ln^{2} (x) + C

\frac{d}{d x} (2 x^{3} - 3 x^{2} - 12 x + 1)

\frac{1}{5 x ^{2} + 6 x}

\frac{\partial}{\partial x} (16 3 y + 2 x)

\frac{\partial}{\partial x} (x e)

\frac{\partial}{\partial y} (y ln (x) + \frac{1}{9} y^{3})

What is the integral of ((ln(x^9)))/x ?
The integral of ((ln(x^9)))/x is 9/2 ln^2(x)+C