What is the integral from 1 to e^2 of (ln^6(x^2))/x ?

The integral from 1 to e^2 of (ln^6(x^2))/x is 8192/7

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integral from 1 to e^2 of (ln^6(x^2))/x

\int_{1}^{e^{2}} \frac{ln ^{6} ( x ^{2} )}{x} d x

\frac{8192}{7}

Decimal Notation

1170.28571 \dots

Solution steps

\int_{1}^{e^{2}} \frac{ln ^{6} ( x ^{2} )}{x} d x

Simplify

\frac{ln ^{6} ( x ^{2} )}{x} : \frac{64 ln ^{6} ( x )}{x}

= \int_{1}^{e^{2}} \frac{64 ln ^{6} ( x )}{x} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 64 \cdot \int_{1}^{e^{2}} \frac{ln ^{6} ( x )}{x} d x

Apply u-substitution

: \int_{0}^{2} u^{6} d u

= 64 \cdot \int_{0}^{2} u^{6} d u

Apply the Power Rule

: [\frac{u ^{7}}{7}]_{0}^{2}

= 64 [\frac{u ^{7}}{7}]_{0}^{2}

Compute the boundaries:

\frac{128}{7}

= 64 \cdot \frac{128}{7}

Simplify

= \frac{8192}{7}

f (x) = \frac{x - 4}{x + 2}

\frac{x}{36 + x ^{2}}

x^{2} y d y - x y^{2} d x - x^{3} y^{2} d x = 0

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

f (x) = (5 x - 1)^{4}, at x = 2

What is the integral from 1 to e^2 of (ln^6(x^2))/x ?
The integral from 1 to e^2 of (ln^6(x^2))/x is 8192/7