integral of (e^a+1)/(e^a)

解

\int \frac{e ^{a} + 1}{e ^{a}} d a

a - \frac{1}{e ^{a}} + C

解答ステップ

\int \frac{e ^{a} + 1}{e ^{a}} d a

拡張

\frac{e ^{a} + 1}{e ^{a}} : 1 + \frac{1}{e ^{a}}

= \int 1 + \frac{1}{e ^{a}} d a

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 a + \int \frac{1}{e ^{a}} d a

\int 1 d a = a

\int \frac{1}{e ^{a}} d a = - \frac{1}{e ^{a}}

= a - \frac{1}{e ^{a}}

定数を解答に追加する

= a - \frac{1}{e ^{a}} + C