What is the integral of (e^{2t}) ?

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

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integral of (e^{2t})

\int (e^{2 t}) d t

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

Solution steps

\int e^{2 t} d t

Apply u-substitution

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

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

Use the common integral:

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

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

Substitute back

u = 2 t

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

Add a constant to the solution

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

\frac{sin ( 3 x )}{tan ( 4 x )}

\frac{1}{x - 1}, 21

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\int_{0}^{\frac{1}{6}} \frac{3}{1 - 9 x ^{2}} d x

\frac{d x}{d t} = x t