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

The integral of e^{8x-9} is 1/8 e^{8x-9}+C

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integral of e^{8x-9}

\int e^{8 x - 9} d x

\frac{1}{8} e^{8 x - 9} + C

Solution steps

\int e^{8 x - 9} d x

Apply u-substitution

: \int e^{u} \frac{1}{8} d u

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

Take the constant out:

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

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

Use the common integral:

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

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

Substitute back

u = 8 x - 9

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

Add a constant to the solution

= \frac{1}{8} e^{8 x - 9} + C

f (x) = ln (x), at x = e^{- 7}

\int \frac{1}{( \frac{1}{3} x - 8 ) ^{5}} d x

\frac{d ^{2}}{d x ^{2}} (3 x^{2} + 4 x)

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

n = 1 \sum \infty \frac{5 n}{2 n + 1}