What is the integral of 3e^{3t} ?

The integral of 3e^{3t} is e^{3t}+C

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integral of 3e^{3t}

\int 3 e^{3 t} d t

e^{3 t} + C

Solution steps

\int 3 e^{3 t} d t

Take the constant out:

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

= 3 \cdot \int e^{3 t} d t

Apply u-substitution

: \int e^{u} \frac{1}{3} d u

= 3 \cdot \int e^{u} \frac{1}{3} d u

Take the constant out:

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

= 3 \cdot \frac{1}{3} \cdot \int e^{u} d u

Use the common integral:

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

= 3 \cdot \frac{1}{3} e^{u}

Substitute back

u = 3 t

= 3 \cdot \frac{1}{3} e^{3 t}

Simplify

3 \cdot \frac{1}{3} e^{3 t} : e^{3 t}

= e^{3 t}

Add a constant to the solution

= e^{3 t} + C

\frac{x}{x + 1}

\int \frac{20 x ^{3} - 40 x ^{2} + 3}{x ^{2} - 2 x} d x

f (x) = \frac{x}{x - 6}

f (x) = 5 x^{2} + 7 x

\frac{d}{d x} (\frac{1}{2 x} - \frac{8}{3 x ^{2}})