What is the integral of 9e^{3t} ?

The integral of 9e^{3t} is 3e^{3t}+C

Popular >

integral of 9e^{3t}

\int 9 e^{3 t} d t

3 e^{3 t} + C

Solution steps

\int 9 e^{3 t} d t

Take the constant out:

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

= 9 \cdot \int e^{3 t} d t

Apply u-substitution

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

= 9 \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

= 9 \cdot \frac{1}{3} \cdot \int e^{u} d u

Use the common integral:

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

= 9 \cdot \frac{1}{3} e^{u}

Substitute back

u = 3 t

= 9 \cdot \frac{1}{3} e^{3 t}

Simplify

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

= 3 e^{3 t}

Add a constant to the solution

= 3 e^{3 t} + C

\frac{d}{d x} (lo g_{5} (2 x - 5))

f (x) = \frac{5}{x}, at x = \frac{1}{4}

\int (y - x) d x

x \to - 1 lim (\frac{x ^{2} - 1}{x - 1})

\int \frac{2 v ^{3} - v}{v ^{4} - v ^{2}} d v