What is the integral of (2x+1)^{17} ?

The integral of (2x+1)^{17} is 1/36 (2x+1)^{18}+C

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integral of (2x+1)^{17}

\int (2 x + 1)^{17} d x

\frac{1}{36} (2 x + 1)^{18} + C

Solution steps

\int (2 x + 1)^{17} d x

Apply u-substitution

: \int \frac{u ^{17}}{2} d u

= \int \frac{u ^{17}}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int u^{17} d u

Apply the Power Rule

: \frac{u ^{18}}{18}

= \frac{1}{2} \cdot \frac{u ^{18}}{18}

Substitute back

u = 2 x + 1

= \frac{1}{2} \cdot \frac{( 2 x + 1 ) ^{18}}{18}

Simplify

\frac{1}{2} \cdot \frac{( 2 x + 1 ) ^{18}}{18} : \frac{1}{36} (2 x + 1)^{18}

= \frac{1}{36} (2 x + 1)^{18}

Add a constant to the solution

= \frac{1}{36} (2 x + 1)^{18} + C

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What is the integral of (2x+1)^{17} ?
The integral of (2x+1)^{17} is 1/36 (2x+1)^{18}+C