integral of 1+3ln(x+1)

解

\int 1 + 3 ln (x + 1) d x

- 2 x + 3 x ln (x + 1) + 3 ln (x + 1) - 3 + C

解答ステップ

\int 1 + 3 ln (x + 1) 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 1 d x + \int 3 ln (x + 1) d x

\int 1 d x = x

\int 3 ln (x + 1) d x = 3 (x ln (x + 1) + ln (x + 1) - x - 1)

= x + 3 (x ln (x + 1) + ln (x + 1) - x - 1)

簡素化

x + 3 (x ln (x + 1) + ln (x + 1) - x - 1) : - 2 x + 3 x ln (x + 1) + 3 ln (x + 1) - 3

= - 2 x + 3 x ln (x + 1) + 3 ln (x + 1) - 3

定数を解答に追加する

= - 2 x + 3 x ln (x + 1) + 3 ln (x + 1) - 3 + C