What is the integral of (e^{3x}) ?

The integral of (e^{3x}) is 1/3 e^{3x}+C

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integral of (e^{3x})

\int (e^{3 x}) d x

\frac{1}{3} e^{3 x} + C

Solution steps

\int e^{3 x} d x

Apply u-substitution

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

= \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

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

Use the common integral:

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

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

Substitute back

u = 3 x

= \frac{1}{3} e^{3 x}

Add a constant to the solution

= \frac{1}{3} e^{3 x} + C

(- 3, 2), (- 1, - 1)

x \to 1 lim (3 x^{2} - 6 x + 1)

x x^{2} + 9

\frac{d}{d x} (x^{4} - 32 x^{2})

y^{^{''}} + 2 y^{^{'}} + 10 y = 0, y (0) = 3, y^{^{'}} (0) = - 1