What is the integral of e^{8x+1} ?

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

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integral of e^{8x+1}

\int e^{8 x + 1} d x

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

Solution steps

\int e^{8 x + 1} d x

Apply u-substitution

= \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 + 1

= \frac{1}{8} e^{8 x + 1}

Add a constant to the solution

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

y = 1 + 3 x, x = 0, x = 8, y = 0

\frac{d}{d x} (\frac{2 x ^{2} - 1}{( x - 1 ) ^{2}})

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

y^{^{'}} = \frac{x}{y ^{2}}

t \frac{d y}{d t} + 2 y = t^{2}