What is the integral of e^x(1+4e^x)^5 ?

The integral of e^x(1+4e^x)^5 is e^x+10e^{2x}+160/3 e^{3x}+160e^{4x}+256e^{5x}+512/3 e^{6x}+C

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integral of e^x(1+4e^x)^5

\int e^{x} (1 + 4 e^{x})^{5} d x

e^{x} + 10 e^{2 x} + \frac{160}{3} e^{3 x} + 160 e^{4 x} + 256 e^{5 x} + \frac{512}{3} e^{6 x} + C

Solution steps

\int e^{x} (1 + 4 e^{x})^{5} d x

Expand

e^{x} (1 + 4 e^{x})^{5} : e^{x} + 20 e^{2 x} + 160 e^{3 x} + 640 e^{4 x} + 1280 e^{5 x} + 1024 e^{6 x}

= \int e^{x} + 20 e^{2 x} + 160 e^{3 x} + 640 e^{4 x} + 1280 e^{5 x} + 1024 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^{x} d x + \int 20 e^{2 x} d x + \int 160 e^{3 x} d x + \int 640 e^{4 x} d x + \int 1280 e^{5 x} d x + \int 1024 e^{6 x} d x

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

\int 20 e^{2 x} d x = 10 e^{2 x}

\int 160 e^{3 x} d x = \frac{160}{3} e^{3 x}

\int 640 e^{4 x} d x = 160 e^{4 x}

\int 1280 e^{5 x} d x = 256 e^{5 x}

\int 1024 e^{6 x} d x = \frac{512}{3} e^{6 x}

= e^{x} + 10 e^{2 x} + \frac{160}{3} e^{3 x} + 160 e^{4 x} + 256 e^{5 x} + \frac{512}{3} e^{6 x}

Add a constant to the solution

= e^{x} + 10 e^{2 x} + \frac{160}{3} e^{3 x} + 160 e^{4 x} + 256 e^{5 x} + \frac{512}{3} e^{6 x} + C

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What is the integral of e^x(1+4e^x)^5 ?
The integral of e^x(1+4e^x)^5 is e^x+10e^{2x}+160/3 e^{3x}+160e^{4x}+256e^{5x}+512/3 e^{6x}+C