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

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

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

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

\frac{1}{4} e^{4 x} - x + C

Solution steps

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

Multiply by the conjugate of

e^{2 x} + 1 : \frac{e ^{4 x} - 1}{e ^{2 x} - 1}

= \int \frac{e ^{4 x} - 1}{e ^{2 x} - 1} (e^{2 x} - 1) d x

Simplify

\frac{e ^{4 x} - 1}{e ^{2 x} - 1} (e^{2 x} - 1) : e^{4 x} - 1

= \int e^{4 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^{4 x} d x - \int 1 d x

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

\int 1 d x = x

= \frac{1}{4} e^{4 x} - x

Add a constant to the solution

= \frac{1}{4} e^{4 x} - x + C

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