integral of (4x^3-2x)ln(3x)

解

\int (4 x^{3} - 2 x) ln (3 x) d x

x^{4} ln (3 x) - \frac{x ^{4}}{4} - x^{2} ln (3 x) + \frac{x ^{2}}{2} + C

解答ステップ

\int (4 x^{3} - 2 x) ln (3 x) d x

拡張

(4 x^{3} - 2 x) ln (3 x) : 4 x^{3} ln (3 x) - 2 x ln (3 x)

= \int 4 x^{3} ln (3 x) - 2 x ln (3 x) d x

総和規則を適用する:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 4 x^{3} ln (3 x) d x - \int 2 x ln (3 x) d x

\int 4 x^{3} ln (3 x) d x = x^{4} ln (3 x) - \frac{x ^{4}}{4}

\int 2 x ln (3 x) d x = x^{2} ln (3 x) - \frac{x ^{2}}{2}

= x^{4} ln (3 x) - \frac{x ^{4}}{4} - (x^{2} ln (3 x) - \frac{x ^{2}}{2})

簡素化

= x^{4} ln (3 x) - \frac{x ^{4}}{4} - x^{2} ln (3 x) + \frac{x ^{2}}{2}

定数を解答に追加する

= x^{4} ln (3 x) - \frac{x ^{4}}{4} - x^{2} ln (3 x) + \frac{x ^{2}}{2} + C