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

解

\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})

解答ステップ

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

定数を除く:

\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

部分積分を適用する

= 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}

限度を計算する:

\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})

簡素化

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