integral of (1-8x)(ln(x)+1)

解

\int (1 - 8 x) (ln (x) + 1) d x

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

解答ステップ

\int (1 - 8 x) (ln (x) + 1) d x

拡張

(1 - 8 x) (ln (x) + 1) : ln (x) + 1 - 8 x ln (x) - 8 x

= \int ln (x) + 1 - 8 x ln (x) - 8 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 ln (x) d x + \int 1 d x - \int 8 x ln (x) d x - \int 8 x d x

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

\int 1 d x = x

\int 8 x ln (x) d x = 8 (\frac{1}{2} x^{2} ln (x) - \frac{x ^{2}}{4})

\int 8 x d x = 4 x^{2}

= x ln (x) - x + x - 8 (\frac{1}{2} x^{2} ln (x) - \frac{x ^{2}}{4}) - 4 x^{2}

簡素化

x ln (x) - x + x - 8 (\frac{1}{2} x^{2} ln (x) - \frac{x ^{2}}{4}) - 4 x^{2} : x ln (x) - 4 x^{2} ln (x) - 2 x^{2}

= x ln (x) - 4 x^{2} ln (x) - 2 x^{2}

定数を解答に追加する

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