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

The integral from 1 to e^3 of (ln^3(x^2))/x is 162

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

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

162

Solution steps

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

Simplify

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

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

Apply u-substitution

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

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

Apply the Power Rule

: [\frac{u ^{4}}{4}]_{0}^{3}

= 8 [\frac{u ^{4}}{4}]_{0}^{3}

Compute the boundaries:

\frac{81}{4}

= 8 \cdot \frac{81}{4}

Simplify

= 162

\frac{x}{1 - x ^{2}}

x \to 0 lim (\frac{( 1 0 ^{x} - 1 )}{x})

x \to 1 + lim (\frac{4}{( x - 1 ) ^{3}})

\frac{\partial}{\partial x} (- 3 cos (2 x))

\int \frac{5}{1 - 9 x ^{2}} + 7 x d x

What is the integral from 1 to e^3 of (ln^3(x^2))/x ?
The integral from 1 to e^3 of (ln^3(x^2))/x is 162