What is the integral of e^{4x}(4+e^{4x})^3 ?

The integral of e^{4x}(4+e^{4x})^3 is 16e^{4x}+6e^{8x}+e^{12x}+1/16 e^{16x}+C

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integral of e^{4x}(4+e^{4x})^3

\int e^{4 x} (4 + e^{4 x})^{3} d x

16 e^{4 x} + 6 e^{8 x} + e^{12 x} + \frac{1}{16} e^{16 x} + C

Solution steps

\int e^{4 x} (4 + e^{4 x})^{3} d x

Expand

e^{4 x} (4 + e^{4 x})^{3} : 64 e^{4 x} + 48 e^{8 x} + 12 e^{12 x} + e^{16 x}

= \int 64 e^{4 x} + 48 e^{8 x} + 12 e^{12 x} + e^{16 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 64 e^{4 x} d x + \int 48 e^{8 x} d x + \int 12 e^{12 x} d x + \int e^{16 x} d x

\int 64 e^{4 x} d x = 16 e^{4 x}

\int 48 e^{8 x} d x = 6 e^{8 x}

\int 12 e^{12 x} d x = e^{12 x}

\int e^{16 x} d x = \frac{1}{16} e^{16 x}

= 16 e^{4 x} + 6 e^{8 x} + e^{12 x} + \frac{1}{16} e^{16 x}

Add a constant to the solution

= 16 e^{4 x} + 6 e^{8 x} + e^{12 x} + \frac{1}{16} e^{16 x} + C

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What is the integral of e^{4x}(4+e^{4x})^3 ?
The integral of e^{4x}(4+e^{4x})^3 is 16e^{4x}+6e^{8x}+e^{12x}+1/16 e^{16x}+C