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

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

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integral of (x^{2n})/(n!)

\int \frac{x ^{2 n}}{n !} d x

\frac{x ^{2 n + 1}}{n ! ( 2 n + 1 )} + C

Solution steps

\int \frac{x ^{2 n}}{n !} d x

Take the constant out:

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

= \frac{1}{n !} \cdot \int x^{2 n} d x

Apply the Power Rule

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

Simplify

\frac{1}{n !} \cdot \frac{x ^{2 n + 1}}{2 n + 1} : \frac{x ^{2 n + 1}}{n ! ( 2 n + 1 )}

= \frac{x ^{2 n + 1}}{n ! ( 2 n + 1 )}

Add a constant to the solution

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

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