What is the integral of (3e^x)/(\sqrt[4]{e^x-1)} ?

The integral of (3e^x)/(\sqrt[4]{e^x-1)} is 4(e^x-1)^{3/4}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (3e^x)/(\sqrt[4]{e^x-1)}

4 (e^{x} - 1)^{\frac{3}{4}} + C

Solution steps

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

Apply u-substitution

Apply the Power Rule

: \frac{4}{3} u^{\frac{3}{4}}

= 3 \cdot \frac{4}{3} u^{\frac{3}{4}}

Substitute back

u = e^{x} - 1

= 3 \cdot \frac{4}{3} (e^{x} - 1)^{\frac{3}{4}}

Simplify

3 \cdot \frac{4}{3} (e^{x} - 1)^{\frac{3}{4}} : 4 (e^{x} - 1)^{\frac{3}{4}}

= 4 (e^{x} - 1)^{\frac{3}{4}}

Add a constant to the solution

= 4 (e^{x} - 1)^{\frac{3}{4}} + C

\int - 10 - 5 x^{2} d x

\int tan^{4} (2 x) sec^{4} (2 x) d x

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

\int \frac{15}{64} + \frac{x}{64} d x

\int \frac{( 3 x - 1 )}{x ^{2} - 2 x + 2} d x

What is the integral of (3e^x)/(\sqrt[4]{e^x-1)} ?
The integral of (3e^x)/(\sqrt[4]{e^x-1)} is 4(e^x-1)^{3/4}+C