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

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

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

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

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

Solution steps

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

Apply u-substitution

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

Apply u-substitution

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

Expand

\frac{( 1 + v ) ^{2}}{v} : \frac{1}{v} + 2 + v

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} (\int \frac{1}{v} d v + \int 2 d v + \int v d v)

\int \frac{1}{v} d v = ln ∣ v ∣

\int 2 d v = 2 v

\int v d v = \frac{v ^{2}}{2}

= \frac{1}{2} (ln ∣ v ∣ + 2 v + \frac{v ^{2}}{2})

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

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