What is the integral of e^{4x-9} ?

The integral of e^{4x-9} is 1/4 e^{4x-9}+C

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integral of e^{4x-9}

\int e^{4 x - 9} d x

\frac{1}{4} e^{4 x - 9} + C

Solution steps

\int e^{4 x - 9} d x

Apply u-substitution

: \int e^{u} \frac{1}{4} d u

= \int e^{u} \frac{1}{4} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int e^{u} d u

Use the common integral:

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

= \frac{1}{4} e^{u}

Substitute back

u = 4 x - 9

= \frac{1}{4} e^{4 x - 9}

Add a constant to the solution

= \frac{1}{4} e^{4 x - 9} + C

\frac{d}{d x} (\frac{3}{e ^{x} + e ^{- x}})

\frac{d}{d x} (\frac{2}{x ^{3} + 1})

y - x y^{^{'}} = a (1 + x^{2} y^{^{'}})

x \to 3 lim (\frac{2 x - 9}{x - 3})

\frac{\partial}{\partial x} (x^{3} y z^{2} + 8 yz)