What is the integral of (e^{3x}-1)/(e^x-1) ?

The integral of (e^{3x}-1)/(e^x-1) is 1/2 e^{2x}+e^x+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^{3x}-1)/(e^x-1)

\int \frac{e ^{3 x} - 1}{e ^{x} - 1} d x

\frac{1}{2} e^{2 x} + e^{x} + x + C

Solution steps

\int \frac{e ^{3 x} - 1}{e ^{x} - 1} d x

Factor

\frac{e ^{3 x} - 1}{e ^{x} - 1}

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

Cancel the common factor:

e^{x} - 1

= \int e^{2 x} + e^{x} + 1 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^{2 x} d x + \int e^{x} d x + \int 1 d x

\int e^{2 x} d x = \frac{1}{2} e^{2 x}

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

\int 1 d x = x

= \frac{1}{2} e^{2 x} + e^{x} + x

Add a constant to the solution

= \frac{1}{2} e^{2 x} + e^{x} + x + C

\int 4 x (1 - x)^{6} d x

\int \frac{2 x - 1}{x ( x - 1 )} d x

\int \frac{x}{( x ^{2} + 5 ) ^{3}} d x

\int x 3 x^{2} - 1 d x

\int ln (5 x + 6) d x

What is the integral of (e^{3x}-1)/(e^x-1) ?
The integral of (e^{3x}-1)/(e^x-1) is 1/2 e^{2x}+e^x+x+C