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

The integral of (e^{3x}-1)^3 is 1/9 e^{9x}-1/2 e^{6x}+e^{3x}-x+C

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

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

\frac{1}{9} e^{9 x} - \frac{1}{2} e^{6 x} + e^{3 x} - x + C

Solution steps

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

Expand

(e^{3 x} - 1)^{3} : e^{9 x} - 3 e^{6 x} + 3 e^{3 x} - 1

= \int e^{9 x} - 3 e^{6 x} + 3 e^{3 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^{9 x} d x - \int 3 e^{6 x} d x + \int 3 e^{3 x} d x - \int 1 d x

\int e^{9 x} d x = \frac{1}{9} e^{9 x}

\int 3 e^{6 x} d x = \frac{1}{2} e^{6 x}

\int 3 e^{3 x} d x = e^{3 x}

\int 1 d x = x

= \frac{1}{9} e^{9 x} - \frac{1}{2} e^{6 x} + e^{3 x} - x

Add a constant to the solution

= \frac{1}{9} e^{9 x} - \frac{1}{2} e^{6 x} + e^{3 x} - x + C

- ln (sin (t))

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

\frac{\partial}{\partial x} (x^{6} y - 3 x^{5} y^{2})

\frac{\partial}{\partial x} (x (3 - 2 x - y))

What is the integral of (e^{3x}-1)^3 ?
The integral of (e^{3x}-1)^3 is 1/9 e^{9x}-1/2 e^{6x}+e^{3x}-x+C