integral of (x+1)[ln(x+1)+c]

解

\int (x + 1) [ln (x + 1) + c] d x

\frac{1}{2} x^{2} ln (x + 1) - \frac{1}{4} x^{2} + \frac{c x ^{2}}{2} + c x - \frac{1}{2} x + x ln (x + 1) + \frac{1}{2} ln (x + 1) - \frac{1}{4} + C

解答ステップ

\int (x + 1) [ln (x + 1) + c] d x

拡張

(x + 1) [ln (x + 1) + c] : x ln (x + 1) + c x + ln (x + 1) + c

= \int x ln (x + 1) + c x + ln (x + 1) + c 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 x ln (x + 1) d x + \int c x d x + \int ln (x + 1) d x + \int c d x

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

\int c x d x = \frac{c x ^{2}}{2}

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

\int c d x = c x

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

簡素化

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

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

定数を解答に追加する

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