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

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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

= 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

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