What is the integral of-e^{4x-8}+1 ?

The integral of-e^{4x-8}+1 is -1/4 e^{4x-8}+x+C

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integral of-e^{4x-8}+1

\int - e^{4 x - 8} + 1 d x

- \frac{1}{4} e^{4 x - 8} + x + C

Solution steps

\int - e^{4 x - 8} + 1 d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \int e^{4 x - 8} d x + \int 1 d x

\int e^{4 x - 8} d x = \frac{1}{4} e^{4 x - 8}

\int 1 d x = x

= - \frac{1}{4} e^{4 x - 8} + x

Add a constant to the solution

= - \frac{1}{4} e^{4 x - 8} + x + C

- 4 e^{x}

f (x) = arcsin (4 x + 1)

\frac{d}{d x} (7 ln (2 x))

y = (\frac{( x - 3 )}{( x + 1 )})^{2}, at x = 0

\frac{x}{x + 1}

What is the integral of-e^{4x-8}+1 ?
The integral of-e^{4x-8}+1 is -1/4 e^{4x-8}+x+C