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

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

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

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

- \frac{4}{3} ln^{3} (x) + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int u^{2} d u

= - 4 \cdot \int u^{2} d u

Apply the Power Rule

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

= - 4 \cdot \frac{u ^{3}}{3}

Substitute back

u = ln (x)

= - 4 \cdot \frac{ln ^{3} ( x )}{3}

Simplify

- 4 \cdot \frac{ln ^{3} ( x )}{3} : - \frac{4}{3} ln^{3} (x)

= - \frac{4}{3} ln^{3} (x)

Add a constant to the solution

= - \frac{4}{3} ln^{3} (x) + C

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