What is the integral of (ln^2(x^4))/x ?

The integral of (ln^2(x^4))/x is 16/3 ln^3(x)+C

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integral of (ln^2(x^4))/x

\int \frac{ln ^{2} ( x ^{4} )}{x} d x

\frac{16}{3} ln^{3} (x) + C

Solution steps

\int \frac{ln ^{2} ( x ^{4} )}{x} d x

Simplify

\frac{ln ^{2} ( x ^{4} )}{x} : \frac{16 ln ^{2} ( x )}{x}

= \int \frac{16 ln ^{2} ( x )}{x} d x

Take the constant out:

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

= 16 \cdot \int \frac{ln ^{2} ( x )}{x} d x

Apply u-substitution

: \int u^{2} d u

= 16 \cdot \int u^{2} d u

Apply the Power Rule

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

= 16 \cdot \frac{u ^{3}}{3}

Substitute back

u = ln (x)

= 16 \cdot \frac{ln ^{3} ( x )}{3}

Simplify

16 \cdot \frac{ln ^{3} ( x )}{3} : \frac{16}{3} ln^{3} (x)

= \frac{16}{3} ln^{3} (x)

Add a constant to the solution

= \frac{16}{3} ln^{3} (x) + C

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What is the integral of (ln^2(x^4))/x ?
The integral of (ln^2(x^4))/x is 16/3 ln^3(x)+C