What is the integral from 1 to e^4 of (ln^3(x^2))/x ?

The integral from 1 to e^4 of (ln^3(x^2))/x is 512

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integral from 1 to e^4 of (ln^3(x^2))/x

\int_{1}^{e^{4}} \frac{ln ^{3} ( x ^{2} )}{x} d x

512

Solution steps

\int_{1}^{e^{4}} \frac{ln ^{3} ( x ^{2} )}{x} d x

Simplify

\frac{ln ^{3} ( x ^{2} )}{x} : \frac{8 ln ^{3} ( x )}{x}

= \int_{1}^{e^{4}} \frac{8 ln ^{3} ( x )}{x} d x

Take the constant out:

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

= 8 \cdot \int_{1}^{e^{4}} \frac{ln ^{3} ( x )}{x} d x

Apply u-substitution

: \int_{0}^{4} u^{3} d u

= 8 \cdot \int_{0}^{4} u^{3} d u

Apply the Power Rule

: [\frac{u ^{4}}{4}]_{0}^{4}

= 8 [\frac{u ^{4}}{4}]_{0}^{4}

Compute the boundaries:

64

= 8 \cdot 64

Simplify

= 512

\frac{d}{d x} (x^{\frac{4}{3}} + 4 x^{\frac{1}{3}})

\frac{\partial}{\partial x} (sin (x^{4} + y^{3}))

f (x) = x^{2} - 3, (4, 13)

x \to 0 + lim (\frac{7}{x} - \frac{7}{∣ x ∣})

\frac{d}{d x} (\frac{2}{1 + x ^{3}})

What is the integral from 1 to e^4 of (ln^3(x^2))/x ?
The integral from 1 to e^4 of (ln^3(x^2))/x is 512