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

The integral of (ln(sqrt(10x)))/x is 1/4 ln^2(10x)+C

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

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

\frac{1}{4} ln^{2} (10 x) + C

Solution steps

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

Apply u-substitution

: \int \frac{ln ( u )}{2 u} d u

= \int \frac{ln ( u )}{2 u} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{ln ( u )}{u} d u

Apply u-substitution

: \int v d v

= \frac{1}{2} \cdot \int v d v

Apply the Power Rule

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

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

Substitute back

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

Simplify

\frac{1}{2} \cdot \frac{ln ^{2} ( 10 x )}{2} : \frac{1}{4} ln^{2} (10 x)

= \frac{1}{4} ln^{2} (10 x)

Add a constant to the solution

= \frac{1}{4} ln^{2} (10 x) + C

y^{^{''}} + 12 y^{^{'}} + 24 y = 4381 e^{2 x} cos (13 x)

\frac{2 s}{( s + 3 ) ^{2}}

\int e^{- s t} sin (b t) d t

y = 1 - x^{2}, y = x^{2} - 1

x^{3} y y^{^{'}} + x^{2} y^{2} = y^{5}

What is the integral of (ln(sqrt(10x)))/x ?
The integral of (ln(sqrt(10x)))/x is 1/4 ln^2(10x)+C