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

The integral from 1 to e^2 of (ln^4(x^2))/x is 512/5

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

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

\frac{512}{5}

Decimal

102.4

Solution steps

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

Simplify

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

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

Apply u-substitution

: \int_{0}^{2} u^{4} d u

= 16 \cdot \int_{0}^{2} u^{4} d u

Apply the Power Rule

: [\frac{u ^{5}}{5}]_{0}^{2}

= 16 [\frac{u ^{5}}{5}]_{0}^{2}

Compute the boundaries:

\frac{32}{5}

= 16 \cdot \frac{32}{5}

Simplify

= \frac{512}{5}

ln (1 - 5 y)

\frac{1}{s ^{2} + 9 s + 9}

f (t) = 2 t^{2} + 3 t - 6

\frac{\partial}{\partial x} (\frac{( a x + b y )}{c x + d y})

x y^{^{''}} + y^{^{'}} = 1

What is the integral from 1 to e^2 of (ln^4(x^2))/x ?
The integral from 1 to e^2 of (ln^4(x^2))/x is 512/5