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

The integral from 1 to e^5 of (ln^3(x^2))/x is 1250

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integral from 1 to e^5 of (ln^3(x^2))/x

\int_{1}^{e^{5}} \frac{ln ^{3} ( x ^{2} )}{x} d x

1250

Solution steps

\int_{1}^{e^{5}} \frac{ln ^{3} ( x ^{2} )}{x} d x

Simplify

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

= \int_{1}^{e^{5}} \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^{5}} \frac{ln ^{3} ( x )}{x} d x

Apply u-substitution

: \int_{0}^{5} u^{3} d u

= 8 \cdot \int_{0}^{5} u^{3} d u

Apply the Power Rule

: [\frac{u ^{4}}{4}]_{0}^{5}

= 8 [\frac{u ^{4}}{4}]_{0}^{5}

Compute the boundaries:

\frac{625}{4}

= 8 \cdot \frac{625}{4}

Simplify

= 1250

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

\frac{\partial}{\partial x} (5 x cos (x y))

\frac{\partial}{\partial x} (- h k^{2} + k^{4})

x y \frac{d y}{d x} = y^{2} + x 4 x^{2} + y^{2}

\frac{\partial}{\partial x} (\frac{- x}{y})

What is the integral from 1 to e^5 of (ln^3(x^2))/x ?
The integral from 1 to e^5 of (ln^3(x^2))/x is 1250