What is the integral of ((-1)^k)/((2k+1)!)(2x)^{2k} ?

The integral of ((-1)^k)/((2k+1)!)(2x)^{2k} is ((-1)^k)/((2k+1)!)*4^k*(x^{2k+1})/(2k+1)+C

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integral of ((-1)^k)/((2k+1)!)(2x)^{2k}

\int \frac{( - 1 ) ^{k}}{( 2 k + 1 ) !} (2 x)^{2 k} d x

\frac{( - 1 ) ^{k}}{( 2 k + 1 ) !} \cdot 4^{k} \cdot \frac{x ^{2 k + 1}}{2 k + 1} + C

Solution steps

\int \frac{( - 1 ) ^{k}}{( 2 k + 1 ) !} (2 x)^{2 k} d x

Take the constant out:

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

= \frac{( - 1 ) ^{k}}{( 2 k + 1 ) !} \cdot \int (2 x)^{2 k} d x

Simplify

(2 x)^{2 k} : 4^{k} x^{2 k}

= \frac{( - 1 ) ^{k}}{( 2 k + 1 ) !} \cdot \int 4^{k} x^{2 k} d x

Take the constant out:

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

= \frac{( - 1 ) ^{k}}{( 2 k + 1 ) !} \cdot 4^{k} \cdot \int x^{2 k} d x

Apply the Power Rule

= \frac{( - 1 ) ^{k}}{( 2 k + 1 ) !} \cdot 4^{k} \cdot \frac{x ^{2 k + 1}}{2 k + 1}

Add a constant to the solution

= \frac{( - 1 ) ^{k}}{( 2 k + 1 ) !} \cdot 4^{k} \cdot \frac{x ^{2 k + 1}}{2 k + 1} + C

\frac{d x}{d t} = x + t

d er i v a t i v e \frac{3}{( 45 + 36 ) ^{3}}

d er i v a t i v e e^{x^{7}}

\frac{d}{d x} (x \frac{2}{3})

\int \frac{1}{x 4 x ^{2} - 9} d x

What is the integral of ((-1)^k)/((2k+1)!)(2x)^{2k} ?
The integral of ((-1)^k)/((2k+1)!)(2x)^{2k} is ((-1)^k)/((2k+1)!)*4^k*(x^{2k+1})/(2k+1)+C