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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

\int (e^{x} + \frac{1}{e ^{x}}) d x

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

Solution steps

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int e^{x} d x = e^{x}

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

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

Add a constant to the solution

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

\int 5 x^{2} e^{5 x} d x

\int \frac{30 x ^{3} - 60 x ^{2} + 8}{x ^{2} - 2 x} d x

\int \frac{x ^{2}}{169 - x ^{2}} d x

\int \frac{x ^{2}}{( x - 6 ) ^{3}} d x

\int tan^{7} (x) sec^{8} (x) d x

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