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

The integral from 1 to e^3 of (ln^4(x^2))/x is 3888/5

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

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

\frac{3888}{5}

Decimal

777.6

Solution steps

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

Simplify

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

= \int_{1}^{e^{3}} \frac{16 ln ^{4} ( 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_{1}^{e^{3}} \frac{ln ^{4} ( x )}{x} d x

Apply u-substitution

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

= 16 \cdot \int_{0}^{3} u^{4} d u

Apply the Power Rule

: [\frac{u ^{5}}{5}]_{0}^{3}

= 16 [\frac{u ^{5}}{5}]_{0}^{3}

Compute the boundaries:

\frac{243}{5}

= 16 \cdot \frac{243}{5}

Simplify

= \frac{3888}{5}

\frac{d}{d x} (3^{s i n (x)})

\frac{d}{d x} (tan (5 x))

x \to 1 lim (3 x^{3} - 5 x^{2} + 5)

\int \frac{x ^{2}}{4 x ^{2} + 1} d x

7 e^{1 - x^{2}}

What is the integral from 1 to e^3 of (ln^4(x^2))/x ?
The integral from 1 to e^3 of (ln^4(x^2))/x is 3888/5