What is the integral of (e^{6x}-1)^3e^{6x} ?

The integral of (e^{6x}-1)^3e^{6x} is 1/24 e^{24x}-1/6 e^{18x}+1/4 e^{12x}-1/6 e^{6x}+C

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integral of (e^{6x}-1)^3e^{6x}

\int (e^{6 x} - 1)^{3} e^{6 x} d x

\frac{1}{24} e^{24 x} - \frac{1}{6} e^{18 x} + \frac{1}{4} e^{12 x} - \frac{1}{6} e^{6 x} + C

Solution steps

\int (e^{6 x} - 1)^{3} e^{6 x} d x

Expand

(e^{6 x} - 1)^{3} e^{6 x} : e^{24 x} - 3 e^{18 x} + 3 e^{12 x} - e^{6 x}

= \int e^{24 x} - 3 e^{18 x} + 3 e^{12 x} - e^{6 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^{24 x} d x - \int 3 e^{18 x} d x + \int 3 e^{12 x} d x - \int e^{6 x} d x

\int e^{24 x} d x = \frac{1}{24} e^{24 x}

\int 3 e^{18 x} d x = \frac{1}{6} e^{18 x}

\int 3 e^{12 x} d x = \frac{1}{4} e^{12 x}

\int e^{6 x} d x = \frac{1}{6} e^{6 x}

= \frac{1}{24} e^{24 x} - \frac{1}{6} e^{18 x} + \frac{1}{4} e^{12 x} - \frac{1}{6} e^{6 x}

Add a constant to the solution

= \frac{1}{24} e^{24 x} - \frac{1}{6} e^{18 x} + \frac{1}{4} e^{12 x} - \frac{1}{6} e^{6 x} + C

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What is the integral of (e^{6x}-1)^3e^{6x} ?
The integral of (e^{6x}-1)^3e^{6x} is 1/24 e^{24x}-1/6 e^{18x}+1/4 e^{12x}-1/6 e^{6x}+C