What is the integral of 3e^{4t} ?

The integral of 3e^{4t} is 3/4 e^{4t}+C

Popular >

integral of 3e^{4t}

\int 3 e^{4 t} d t

\frac{3}{4} e^{4 t} + C

Solution steps

\int 3 e^{4 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^{4 t} d t

Apply u-substitution

: \int e^{u} \frac{1}{4} d u

= 3 \cdot \int e^{u} \frac{1}{4} d u

Take the constant out:

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

= 3 \cdot \frac{1}{4} \cdot \int e^{u} d u

Use the common integral:

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

= 3 \cdot \frac{1}{4} e^{u}

Substitute back

u = 4 t

= 3 \cdot \frac{1}{4} e^{4 t}

Simplify

3 \cdot \frac{1}{4} e^{4 t} : \frac{3}{4} e^{4 t}

= \frac{3}{4} e^{4 t}

Add a constant to the solution

= \frac{3}{4} e^{4 t} + C

\int \frac{4 + 8 x sin ( x )}{2 x} d x

\int tan^{8} (x) cos^{9} (x) d x

y = 1 0^{x}

(\frac{1}{3}) (\frac{d x}{d t}) - 2 e^{- t} = - x

\int_{0}^{1} x (1 - x)^{2} d x