What is the integral of (e^a+1)/(e^a) ?

The integral of (e^a+1)/(e^a) is a-1/(e^a)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

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

Expand

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

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

Apply the Sum Rule:

\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}}

Add a constant to the solution

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

x \to - 8 lim (\frac{x ^{2} - 4}{8 - x})

\int \frac{1}{4 e ^{- 4 x} + e ^{4 x}} d x

x \to 0 lim (f (x) (x^{2}))

\frac{d x}{d t} = 1

y = x^{2} - x^{4}, (0, 1)

What is the integral of (e^a+1)/(e^a) ?
The integral of (e^a+1)/(e^a) is a-1/(e^a)+C