What is the integral of (-x)^n ?

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

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integral of (-x)^n

\int (- x)^{n} d x

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

Solution steps

\int (- x)^{n} d x

Apply u-substitution

: \int - u^{n} d u

= \int - u^{n} d u

Take the constant out:

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

= - \int u^{n} d u

Apply the Power Rule

: \frac{u ^{n + 1}}{n + 1}

= - \frac{u ^{n + 1}}{n + 1}

Substitute back

u = - x

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

Add a constant to the solution

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

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