What is the integral of (ln(19x))/x ?

The integral of (ln(19x))/x is (ln^2(19x))/2+C

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integral of (ln(19x))/x

\int \frac{ln ( 19 x )}{x} d x

\frac{ln ^{2} ( 19 x )}{2} + C

Solution steps

\int \frac{ln ( 19 x )}{x} d x

Apply u-substitution

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

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

Apply u-substitution

: \int v d v

= \int v d v

Apply the Power Rule

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

= \frac{v ^{2}}{2}

Substitute back

= \frac{ln ^{2} ( 19 x )}{2}

Add a constant to the solution

= \frac{ln ^{2} ( 19 x )}{2} + C

\frac{\partial}{\partial x} (5 x^{2} y - 1)

f (x) = \frac{- 4 x}{x ^{2} + 1}, (1, 2)

\int \frac{3}{2 x - 1} d x

y^{^{''}} + 4 y = 8 t^{2} - 20 t + 8

\frac{d}{d x} (4 x^{3} + \frac{1}{x ^{2}} - x + 3)

What is the integral of (ln(19x))/x ?
The integral of (ln(19x))/x is (ln^2(19x))/2+C