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

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

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integral of e^{9-4x}

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

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

Solution steps

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

Apply u-substitution

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

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

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

Add a constant to the solution

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

\int \frac{cos ( x )}{3 + 2 sin ( x )} d x

y = 6 x (x^{2} + 2) 2

\int \frac{( 3 x + 11 )}{x ^{2} - x - 6} d x

x + 11

f (x) = \frac{4 x}{x + 2}, at x = 2