What is the integral of (-1)^{x+1} ?

The integral of (-1)^{x+1} is -((-1)^x)/(ln(-1))+C

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integral of (-1)^{x+1}

\int (- 1)^{x + 1} d x

- \frac{( - 1 ) ^{x}}{ln ( - 1 )} + C

Solution steps

\int (- 1)^{x + 1} d x

(- 1)^{x + 1} = (- 1)^{1} (- 1)^{x}

= \int (- 1)^{1} (- 1)^{x} d x

Take the constant out:

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

= (- 1)^{1} \cdot \int (- 1)^{x} d x

Simplify

= (- 1) \cdot \int (- 1)^{x} d x

\int a^{x} d x = \frac{a ^{x}}{ln a}

= (- 1) \frac{( - 1 ) ^{x}}{ln ( - 1 )}

Simplify

= - \frac{( - 1 ) ^{x}}{ln ( - 1 )}

Add a constant to the solution

= - \frac{( - 1 ) ^{x}}{ln ( - 1 )} + C

\int \frac{4}{3 x - 6} d x

\int \frac{x ^{2} + 8 x + 9}{( x - 1 ) ( x + 2 ) ^{2}} d x

\int \frac{4}{x ^{\frac{1}{2}}} d x

\int \frac{x + 8}{x ^{2} + x - 6} d x

\int \frac{- 4 x + 8}{( x ^{2} + 1 ) ( x - 1 )} d x

What is the integral of (-1)^{x+1} ?
The integral of (-1)^{x+1} is -((-1)^x)/(ln(-1))+C