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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply the Power Rule

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

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

Simplify

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

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

Add a constant to the solution

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

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