What is the integral of 8e^{9x} ?

The integral of 8e^{9x} is 8/9 e^{9x}+C

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integral of 8e^{9x}

\int 8 e^{9 x} d x

\frac{8}{9} e^{9 x} + C

Solution steps

\int 8 e^{9 x} d x

Take the constant out:

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

= 8 \cdot \int e^{9 x} d x

Apply u-substitution

: \int e^{u} \frac{1}{9} d u

= 8 \cdot \int e^{u} \frac{1}{9} d u

Take the constant out:

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

= 8 \cdot \frac{1}{9} \cdot \int e^{u} d u

Use the common integral:

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

= 8 \cdot \frac{1}{9} e^{u}

Substitute back

u = 9 x

= 8 \cdot \frac{1}{9} e^{9 x}

Simplify

8 \cdot \frac{1}{9} e^{9 x} : \frac{8}{9} e^{9 x}

= \frac{8}{9} e^{9 x}

Add a constant to the solution

= \frac{8}{9} e^{9 x} + C

y^{^{'''}} - 5 y^{^{''}} + 6 y^{^{'}} = 0

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

\frac{\partial}{\partial x} (3 x e^{2 x y})

\int \frac{( 8 x ^{2} + 3 x + 8 )}{( x ^{2} + 1 ) ^{2}} d x

y^{^{'}} = x e^{x}