What is the integral of 8e^{4x}(e^{4x}-4)^{1/5} ?

The integral of 8e^{4x}(e^{4x}-4)^{1/5} is 5/3 (e^{4x}-4)^{6/5}+C

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integral of 8e^{4x}(e^{4x}-4)^{1/5}

\int 8 e^{4 x} (e^{4 x} - 4)^{\frac{1}{5}} d x

\frac{5}{3} (e^{4 x} - 4)^{\frac{6}{5}} + C

Solution steps

\int 8 e^{4 x} (e^{4 x} - 4)^{\frac{1}{5}} d x

Take the constant out:

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

= 8 \cdot \int e^{4 x} (e^{4 x} - 4)^{\frac{1}{5}} d x

Apply u-substitution

= 8 \cdot \int \frac{u ^{\frac{1}{5}}}{4} d u

Take the constant out:

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

= 8 \cdot \frac{1}{4} \cdot \int u^{\frac{1}{5}} d u

Apply the Power Rule

= 8 \cdot \frac{1}{4} \cdot \frac{5}{6} u^{\frac{6}{5}}

Substitute back

u = e^{4 x} - 4

= 8 \cdot \frac{1}{4} \cdot \frac{5}{6} (e^{4 x} - 4)^{\frac{6}{5}}

Simplify

8 \cdot \frac{1}{4} \cdot \frac{5}{6} (e^{4 x} - 4)^{\frac{6}{5}} : \frac{5}{3} (e^{4 x} - 4)^{\frac{6}{5}}

= \frac{5}{3} (e^{4 x} - 4)^{\frac{6}{5}}

Add a constant to the solution

= \frac{5}{3} (e^{4 x} - 4)^{\frac{6}{5}} + C

What is the integral of 8e^{4x}(e^{4x}-4)^{1/5} ?
The integral of 8e^{4x}(e^{4x}-4)^{1/5} is 5/3 (e^{4x}-4)^{6/5}+C