What is the integral of (x+1)^{15} ?

The integral of (x+1)^{15} is ((x+1)^{16})/(16)+C

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integral of (x+1)^{15}

\int (x + 1)^{15} d x

\frac{( x + 1 ) ^{16}}{16} + C

Solution steps

\int (x + 1)^{15} d x

Apply u-substitution

: \int u^{15} d u

= \int u^{15} d u

Apply the Power Rule

: \frac{u ^{16}}{16}

= \frac{u ^{16}}{16}

Substitute back

u = x + 1

= \frac{( x + 1 ) ^{16}}{16}

Add a constant to the solution

= \frac{( x + 1 ) ^{16}}{16} + C

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What is the integral of (x+1)^{15} ?
The integral of (x+1)^{15} is ((x+1)^{16})/(16)+C