What is the integral from a to b of 4x^5(ln(x^3)) ?

The integral from a to b of 4x^5(ln(x^3)) is 4((b^6ln(b)-a^6ln(a))/2+(-b^6+a^6)/(12))

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integral from a to b of 4x^5(ln(x^3))

\int_{a}^{b} 4 x^{5} (ln (x^{3})) d x

4 (\frac{b ^{6} ln ( b ) - a ^{6} ln ( a )}{2} + \frac{- b ^{6} + a ^{6}}{12})

Solution steps

\int_{a}^{b} 4 x^{5} ln (x^{3}) d x

Take the constant out:

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

= 4 \cdot \int_{a}^{b} x^{5} ln (x^{3}) d x

Apply Integration By Parts

: [\frac{1}{6} x^{6} ln (x^{3}) - \int \frac{x ^{5}}{2} d x]_{a}^{b}

= 4 [\frac{1}{6} x^{6} ln (x^{3}) - \int \frac{x ^{5}}{2} d x]_{a}^{b}

\int \frac{x ^{5}}{2} d x = \frac{x ^{6}}{12}

= 4 [\frac{1}{6} x^{6} ln (x^{3}) - \frac{x ^{6}}{12}]_{a}^{b}

\frac{1}{6} x^{6} ln (x^{3}) = \frac{1}{2} x^{6} ln (x)

= 4 [\frac{1}{2} x^{6} ln (x) - \frac{x ^{6}}{12}]_{a}^{b}

Compute the boundaries:

\frac{1}{2} b^{6} ln (b) - \frac{b ^{6}}{12} - \frac{1}{2} a^{6} ln (a) + \frac{a ^{6}}{12}

= 4 (\frac{1}{2} b^{6} ln (b) - \frac{b ^{6}}{12} - \frac{1}{2} a^{6} ln (a) + \frac{a ^{6}}{12})

Simplify

= 4 (\frac{b ^{6} ln ( b ) - a ^{6} ln ( a )}{2} + \frac{- b ^{6} + a ^{6}}{12})

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What is the integral from a to b of 4x^5(ln(x^3)) ?
The integral from a to b of 4x^5(ln(x^3)) is 4((b^6ln(b)-a^6ln(a))/2+(-b^6+a^6)/(12))