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

The integral of e^{3x+4} is 1/3 e^{3x+4}+C

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integral of e^{3x+4}

\int e^{3 x + 4} d x

\frac{1}{3} e^{3 x + 4} + C

Solution steps

\int e^{3 x + 4} 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 + 4

= \frac{1}{3} e^{3 x + 4}

Add a constant to the solution

= \frac{1}{3} e^{3 x + 4} + C

x \to 2 - lim (- 2)

y = x + 3, y = \frac{1}{2} x + 3

10 y^{^{''}} - y^{^{'}} - 3 y = 0

\frac{\partial}{\partial x} (5 x^{2} + 5 y^{2})

\frac{\partial}{\partial x} (x \cdot e^{x - y - z})