What is the integral of (ln(x^{1/2}))/x ?

The integral of (ln(x^{1/2}))/x is 1/4 ln^2(x)+C

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integral of (ln(x^{1/2}))/x

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

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

Solution steps

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

Simplify

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

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

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 ( x )}{x} d x

Apply u-substitution

: \int u d u

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

Apply the Power Rule

: \frac{u ^{2}}{2}

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

Substitute back

u = ln (x)

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

Simplify

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

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

Add a constant to the solution

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

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