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

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

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integral of (ln(x^2))/(2x)

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

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

Solution steps

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

Apply u-substitution

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

Apply u-substitution

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

Apply the Power Rule

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

\frac{d y}{d x} = \frac{1}{10} y - \frac{1}{5000} y^{2}, y (0) = 50

x \to π lim (1)

d er i v a t i v e f (x) = \frac{e ^{5 x}}{2 x - 5}

x \to 1 lim (\frac{1 - x ^{2}}{ln ( x )})

\frac{d}{d x} (2 e^{- 1.5 x} cos (3 x))

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