What is the integral of (ln(x^{28}))/x ?

The integral of (ln(x^{28}))/x is 14ln^2(x)+C

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integral of (ln(x^{28}))/x

\int \frac{ln ( x ^{28} )}{x} d x

14 ln^{2} (x) + C

Solution steps

\int \frac{ln ( x ^{28} )}{x} d x

Apply u-substitution

: \int \frac{ln ( u )}{28 u} d u

= \int \frac{ln ( u )}{28 u} d u

Take the constant out:

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

= \frac{1}{28} \cdot \int \frac{ln ( u )}{u} d u

Apply u-substitution

: \int v d v

= \frac{1}{28} \cdot \int v d v

Apply the Power Rule

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

= \frac{1}{28} \cdot \frac{v ^{2}}{2}

Substitute back

= \frac{1}{28} \cdot \frac{ln ^{2} ( x ^{28} )}{2}

Simplify

\frac{1}{28} \cdot \frac{ln ^{2} ( x ^{28} )}{2} : 14 ln^{2} (x)

= 14 ln^{2} (x)

Add a constant to the solution

= 14 ln^{2} (x) + C

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

\frac{\partial}{\partial x} (4 x (x^{2} + y^{2}) - 4 a^{2} x)

\frac{a}{x}

\frac{s}{( s - 1 )}

y (1 + x^{2}) y^{^{'}} - x (6 + y^{2}) = 0

What is the integral of (ln(x^{28}))/x ?
The integral of (ln(x^{28}))/x is 14ln^2(x)+C