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

The integral from 1 to e^2 of (ln^3(x^2))/x is 32

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

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

32

Solution steps

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

Simplify

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

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

Apply u-substitution

: \int_{0}^{2} u^{3} d u

= 8 \cdot \int_{0}^{2} u^{3} d u

Apply the Power Rule

: [\frac{u ^{4}}{4}]_{0}^{2}

= 8 [\frac{u ^{4}}{4}]_{0}^{2}

Compute the boundaries:

4

= 8 \cdot 4

Simplify

= 32

n = 0 \sum \infty \frac{n !}{1 0 ^{n}}

\frac{d y}{d x} = y (8 x^{2} + 6)

\int_{e}^{\infty} \frac{ln ( x )}{x} d x

f (x) = \frac{8 x}{x ^{2} + 1}

(4 - t^{2}) y^{^{'}} + 2 t y = 3 t^{2}

What is the integral from 1 to e^2 of (ln^3(x^2))/x ?
The integral from 1 to e^2 of (ln^3(x^2))/x is 32