What is the integral of e^{8x+6} ?

The integral of e^{8x+6} is 1/8 e^{8x+6}+C

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integral of e^{8x+6}

\int e^{8 x + 6} d x

\frac{1}{8} e^{8 x + 6} + C

Solution steps

\int e^{8 x + 6} 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 + 6

= \frac{1}{8} e^{8 x + 6}

Add a constant to the solution

= \frac{1}{8} e^{8 x + 6} + C

y^{^{'}} = \frac{( cos ( 18 x ) )}{( e ^{(18 \cdot y)} )}

x^{2} y^{^{''}} + 3 x y^{^{'}} - 3 y = 0

d y - (y - 9)^{2} d x = 0

x^{\frac{5}{6}}

\frac{x ( 2 - x )}{e ^{x}}