What is the integral of e^{2x}(e^{2x}+1)^4 ?

The integral of e^{2x}(e^{2x}+1)^4 is 1/10 e^{10x}+1/2 e^{8x}+e^{6x}+e^{4x}+1/2 e^{2x}+C

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integral of e^{2x}(e^{2x}+1)^4

\int e^{2 x} (e^{2 x} + 1)^{4} d x

\frac{1}{10} e^{10 x} + \frac{1}{2} e^{8 x} + e^{6 x} + e^{4 x} + \frac{1}{2} e^{2 x} + C

Solution steps

\int e^{2 x} (e^{2 x} + 1)^{4} d x

Expand

e^{2 x} (e^{2 x} + 1)^{4} : e^{10 x} + 4 e^{8 x} + 6 e^{6 x} + 4 e^{4 x} + e^{2 x}

= \int e^{10 x} + 4 e^{8 x} + 6 e^{6 x} + 4 e^{4 x} + e^{2 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^{10 x} d x + \int 4 e^{8 x} d x + \int 6 e^{6 x} d x + \int 4 e^{4 x} d x + \int e^{2 x} d x

\int e^{10 x} d x = \frac{1}{10} e^{10 x}

\int 4 e^{8 x} d x = \frac{1}{2} e^{8 x}

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

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

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

= \frac{1}{10} e^{10 x} + \frac{1}{2} e^{8 x} + e^{6 x} + e^{4 x} + \frac{1}{2} e^{2 x}

Add a constant to the solution

= \frac{1}{10} e^{10 x} + \frac{1}{2} e^{8 x} + e^{6 x} + e^{4 x} + \frac{1}{2} e^{2 x} + C

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What is the integral of e^{2x}(e^{2x}+1)^4 ?
The integral of e^{2x}(e^{2x}+1)^4 is 1/10 e^{10x}+1/2 e^{8x}+e^{6x}+e^{4x}+1/2 e^{2x}+C