What is the integral of 3ye^{3x} ?

The integral of 3ye^{3x} is ye^{3x}+C

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integral of 3ye^{3x}

\int 3 y e^{3 x} d x

y e^{3 x} + C

Solution steps

\int 3 y e^{3 x} d x

Take the constant out:

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

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

Apply u-substitution

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

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

Use the common integral:

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

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

Substitute back

u = 3 x

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

Simplify

3 y \frac{1}{3} e^{3 x} : y e^{3 x}

= y e^{3 x}

Add a constant to the solution

= y e^{3 x} + C

f (x) = \frac{2}{3 x + 5}, at x = - 1

4 y^{^{''}} - 4 y^{^{'}} + 50 y = 0

y = 8^{x}

\int \frac{1 - 2 x - x ^{2}}{x ^{3} + x ^{2} - 3 x + 1} d x

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