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

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

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

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

\frac{1}{2} ln e^{4 x} + 1 + e^{2 x} + C

Solution steps

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

Apply u-substitution

: \int \frac{e ^{u}}{2 e ^{2 u} + 1} d u

= \int \frac{e ^{u}}{2 e ^{2 u} + 1} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int \frac{e ^{u}}{e ^{2 u} + 1} d u

Apply u-substitution

: \int \frac{1}{v ^{2} + 1} d v

= \frac{1}{2} \cdot \int \frac{1}{v ^{2} + 1} d v

Use the common integral:

\int \frac{1}{v ^{2} + 1} d v = ln v^{2} + 1 + v

= \frac{1}{2} ln v^{2} + 1 + v

Substitute back

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

(e^{2 x})^{2} = e^{4 x}

= \frac{1}{2} ln e^{4 x} + 1 + e^{2 x}

Add a constant to the solution

= \frac{1}{2} ln e^{4 x} + 1 + e^{2 x} + C

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