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

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

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integral of e^{4-3x}

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

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

Solution steps

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

Apply u-substitution

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

= \int - \frac{1}{3} e^{u} 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 = 4 - 3 x

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

Add a constant to the solution

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

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

\frac{s + 3}{( s + 1 ) ( s + 2 )}

\frac{\partial}{\partial x} (- y cos (x))

y^{^{''}} + 4 y^{^{'}} + 3 y = 4 e^{- x}

\frac{\partial}{\partial x} (x \cdot sin (x + y))