What is the integral of (ln^2(x))/(3x) ?

The integral of (ln^2(x))/(3x) is 1/9 ln^3(x)+C

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integral of (ln^2(x))/(3x)

\int \frac{ln ^{2} ( x )}{3 x} d x

\frac{1}{9} ln^{3} (x) + C

Solution steps

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

Apply u-substitution

: \int u^{2} d u

= \frac{1}{3} \cdot \int u^{2} d u

Apply the Power Rule

: \frac{u ^{3}}{3}

= \frac{1}{3} \cdot \frac{u ^{3}}{3}

Substitute back

u = ln (x)

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

Simplify

\frac{1}{3} \cdot \frac{ln ^{3} ( x )}{3} : \frac{1}{9} ln^{3} (x)

= \frac{1}{9} ln^{3} (x)

Add a constant to the solution

= \frac{1}{9} ln^{3} (x) + C

\int \frac{6 x + 5}{e ^{x}} d x

y = ln (1 - x^{2})

\frac{\partial}{\partial y} (2 y^{3} - 6 y + x^{2})

ln (5)

x \to + 0 + lim (\frac{1}{x} - \frac{1}{x ^{2} + x})

What is the integral of (ln^2(x))/(3x) ?
The integral of (ln^2(x))/(3x) is 1/9 ln^3(x)+C