integral of (4x^3-3x^2+1)ln(x)

解

\int (4 x^{3} - 3 x^{2} + 1) ln (x) d x

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

解答ステップ

\int (4 x^{3} - 3 x^{2} + 1) ln (x) d x

拡張

(4 x^{3} - 3 x^{2} + 1) ln (x) : 4 x^{3} ln (x) - 3 x^{2} ln (x) + ln (x)

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

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

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

\int ln (x) d x = x ln (x) - x

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

簡素化

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

定数を解答に追加する

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