What is the integral of (ln(10x))/(3x) ?

The integral of (ln(10x))/(3x) is 1/6 ln^2(10x)+C

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integral of (ln(10x))/(3x)

\int \frac{ln ( 10 x )}{3 x} d x

\frac{1}{6} ln^{2} (10 x) + C

Solution steps

\int \frac{ln ( 10 x )}{3 x} d x

Take the constant out:

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

= \frac{1}{3} \cdot \int \frac{ln ( 10 x )}{x} d x

Apply u-substitution

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

= \frac{1}{3} \cdot \int \frac{ln ( u )}{u} d u

Apply u-substitution

: \int v d v

= \frac{1}{3} \cdot \int v d v

Apply the Power Rule

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

= \frac{1}{3} \cdot \frac{v ^{2}}{2}

Substitute back

= \frac{1}{3} \cdot \frac{ln ^{2} ( 10 x )}{2}

Simplify

\frac{1}{3} \cdot \frac{ln ^{2} ( 10 x )}{2} : \frac{1}{6} ln^{2} (10 x)

= \frac{1}{6} ln^{2} (10 x)

Add a constant to the solution

= \frac{1}{6} ln^{2} (10 x) + C

\int - 2 x e^{- 2 x} d x

\frac{1}{2} t^{2} - 6 t + 23

\int (x^{3} ln (x)) d x

\int \frac{a ^{x}}{x} d x

\frac{d}{d θ} (ln (θ + 1))

What is the integral of (ln(10x))/(3x) ?
The integral of (ln(10x))/(3x) is 1/6 ln^2(10x)+C