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

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

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

\int \frac{e ^{2 x}}{e ^{2 x} + 4 e ^{x} + 2} d x

\frac{1}{2 2} e^{x} ln \frac{1}{2} (e^{x} + 2) - 1 - \frac{1}{2 2} e^{x} ln \frac{1}{2} e^{x} + 2 - 1 + \frac{2 + 1}{2} ln \frac{1}{2} (e^{x} + 2) + 1 + \frac{1 - 2}{2} ln \frac{1}{2} e^{x} + 2 - 1 - 1 + C

Solution steps

\int \frac{e ^{2 x}}{e ^{2 x} + 4 e ^{x} + 2} d x

Apply u-substitution

: \int \frac{u}{u ^{2} + 4 u + 2} d u

= \int \frac{u}{u ^{2} + 4 u + 2} d u

Apply Integration By Parts

: - \frac{1}{2 2} u (ln \frac{1}{2} (u + 2) + 1 - ln \frac{1}{2} (u + 2) - 1) - \int - \frac{1}{2 2} (ln \frac{1}{2} (u + 2) + 1 - ln \frac{1}{2} (u + 2) - 1) d u

= - \frac{1}{2 2} u (ln \frac{1}{2} (u + 2) + 1 - ln \frac{1}{2} (u + 2) - 1) - \int - \frac{1}{2 2} (ln \frac{1}{2} (u + 2) + 1 - ln \frac{1}{2} (u + 2) - 1) d u

\int - \frac{1}{2 2} (ln \frac{1}{2} (u + 2) + 1 - ln \frac{1}{2} (u + 2) - 1) d u = - \frac{1}{2 2} (u ln \frac{1}{2} (u + 2) + 1 - u ln \frac{1}{2} u + 2 - 1 + 2 ln \frac{1}{2} (u + 2) + 1 + 2 ln \frac{1}{2} (u + 2) + 1 + 2 ln \frac{u}{2} + 2 - 1 - 2 ln \frac{u}{2} + 2 - 1 - 22)

= - \frac{1}{2 2} u (ln \frac{1}{2} (u + 2) + 1 - ln \frac{1}{2} (u + 2) - 1) - (- \frac{1}{2 2} (u ln \frac{1}{2} (u + 2) + 1 - u ln \frac{1}{2} u + 2 - 1 + 2 ln \frac{1}{2} (u + 2) + 1 + 2 ln \frac{1}{2} (u + 2) + 1 + 2 ln \frac{u}{2} + 2 - 1 - 2 ln \frac{u}{2} + 2 - 1 - 22))

Substitute back

u = e^{x}

= - \frac{1}{2 2} e^{x} (ln \frac{1}{2} (e^{x} + 2) + 1 - ln \frac{1}{2} (e^{x} + 2) - 1) - (- \frac{1}{2 2} (e^{x} ln \frac{1}{2} (e^{x} + 2) + 1 - e^{x} ln \frac{1}{2} e^{x} + 2 - 1 + 2 ln \frac{1}{2} (e^{x} + 2) + 1 + 2 ln \frac{1}{2} (e^{x} + 2) + 1 + 2 ln \frac{e ^{x}}{2} + 2 - 1 - 2 ln \frac{e ^{x}}{2} + 2 - 1 - 22))

Simplify

- \frac{1}{2 2} e^{x} (ln \frac{1}{2} (e^{x} + 2) + 1 - ln \frac{1}{2} (e^{x} + 2) - 1) - (- \frac{1}{2 2} (e^{x} ln \frac{1}{2} (e^{x} + 2) + 1 - e^{x} ln \frac{1}{2} e^{x} + 2 - 1 + 2 ln \frac{1}{2} (e^{x} + 2) + 1 + 2 ln \frac{1}{2} (e^{x} + 2) + 1 + 2 ln \frac{e ^{x}}{2} + 2 - 1 - 2 ln \frac{e ^{x}}{2} + 2 - 1 - 22)) : \frac{1}{2 2} e^{x} ln \frac{1}{2} (e^{x} + 2) - 1 - \frac{1}{2 2} e^{x} ln \frac{1}{2} e^{x} + 2 - 1 + \frac{2 + 1}{2} ln \frac{1}{2} (e^{x} + 2) + 1 + \frac{1 - 2}{2} ln \frac{1}{2} e^{x} + 2 - 1 - 1

= \frac{1}{2 2} e^{x} ln \frac{1}{2} (e^{x} + 2) - 1 - \frac{1}{2 2} e^{x} ln \frac{1}{2} e^{x} + 2 - 1 + \frac{2 + 1}{2} ln \frac{1}{2} (e^{x} + 2) + 1 + \frac{1 - 2}{2} ln \frac{1}{2} e^{x} + 2 - 1 - 1

Add a constant to the solution

= \frac{1}{2 2} e^{x} ln \frac{1}{2} (e^{x} + 2) - 1 - \frac{1}{2 2} e^{x} ln \frac{1}{2} e^{x} + 2 - 1 + \frac{2 + 1}{2} ln \frac{1}{2} (e^{x} + 2) + 1 + \frac{1 - 2}{2} ln \frac{1}{2} e^{x} + 2 - 1 - 1 + C

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