What is the integral of 6e^{3x} ?

The integral of 6e^{3x} is 2e^{3x}+C

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integral of 6e^{3x}

\int 6 e^{3 x} d x

2 e^{3 x} + C

Solution steps

\int 6 e^{3 x} d x

Take the constant out:

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

= 6 \cdot \int e^{3 x} d x

Apply u-substitution

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

= 6 \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

= 6 \cdot \frac{1}{3} \cdot \int e^{u} d u

Use the common integral:

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

= 6 \cdot \frac{1}{3} e^{u}

Substitute back

u = 3 x

= 6 \cdot \frac{1}{3} e^{3 x}

Simplify

6 \cdot \frac{1}{3} e^{3 x} : 2 e^{3 x}

= 2 e^{3 x}

Add a constant to the solution

= 2 e^{3 x} + C

x \to 0 lim (\frac{sin ( 3 x )}{3})

x \to 0 lim (\frac{x ^{4} + 2 x}{x})

\frac{d y}{d x} = \frac{( x ^{2} + 3 y ^{2} )}{( 2 x y )}

\int \frac{1}{x ^{2} + a} d x

\frac{d y}{d x} = 2 + 3 + y - 2 x