What is the integral of (x^n)/(n+1) ?

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

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integral of (x^n)/(n+1)

\int \frac{x ^{n}}{n + 1} d x

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

Solution steps

\int \frac{x ^{n}}{n + 1} d x

Take the constant out:

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

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

Apply the Power Rule

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

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

Simplify

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

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

Add a constant to the solution

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

(78) (7 - 9)

e^{- t \cdot 3}

\int \frac{16}{x} d x

\int_{- 2}^{2} \frac{1}{2} (4 - x^{2}) d x

\frac{3}{5 x + 2}, at x = - 1

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