What is the integral of (-x)^{2n} ?

The integral of (-x)^{2n} is (x^{2n+1})/(2n+1)+C

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integral of (-x)^{2n}

\int (- x)^{2 n} d x

\frac{x ^{2 n + 1}}{2 n + 1} + C

Solution steps

\int (- x)^{2 n} d x

Apply exponent rule:

(a \cdot b)^{n} = a^{n} \cdot b^{n}

(- x)^{2 n} = (- 1)^{2 n} (x)^{2 n}

= \int (- 1)^{2 n} x^{2 n} d x

Simplify

(- 1)^{2 n} x^{2 n} : x^{2 n}

= \int x^{2 n} d x

Apply the Power Rule

: \frac{x ^{2 n + 1}}{2 n + 1}

= \frac{x ^{2 n + 1}}{2 n + 1}

Add a constant to the solution

= \frac{x ^{2 n + 1}}{2 n + 1} + C

f (x) = 2 t^{- \frac{3}{4}}

\int t e^{t^{2}} d t

\frac{d}{d x} (\frac{3 x ^{2} - 5 x - 2}{4 x + 12})

\frac{\partial}{\partial y} (x^{3} - 3 x + y - e^{y})

x \to 9 lim (\frac{x ^{2} - 2}{x - 9})

What is the integral of (-x)^{2n} ?
The integral of (-x)^{2n} is (x^{2n+1})/(2n+1)+C