What is the integral of e^{2u} ?

The integral of e^{2u} is 1/2 e^{2u}+C

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integral of e^{2u}

\int e^{2 u} d u

\frac{1}{2} e^{2 u} + C

Solution steps

\int e^{2 u} d u

Apply u-substitution

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

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

Take the constant out:

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

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

Use the common integral:

\int e^{v} d v = e^{v}

= \frac{1}{2} e^{v}

Substitute back

v = 2 u

= \frac{1}{2} e^{2 u}

Add a constant to the solution

= \frac{1}{2} e^{2 u} + C

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