What is the integral of (2x+1)^{20} ?

The integral of (2x+1)^{20} is 1/42 (2x+1)^{21}+C

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integral of (2x+1)^{20}

\int (2 x + 1)^{20} d x

\frac{1}{42} (2 x + 1)^{21} + C

Solution steps

\int (2 x + 1)^{20} d x

Apply u-substitution

= \int \frac{u ^{20}}{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^{20} d u

Apply the Power Rule

= \frac{1}{2} \cdot \frac{u ^{21}}{21}

Substitute back

u = 2 x + 1

= \frac{1}{2} \cdot \frac{( 2 x + 1 ) ^{21}}{21}

Simplify

\frac{1}{2} \cdot \frac{( 2 x + 1 ) ^{21}}{21} : \frac{1}{42} (2 x + 1)^{21}

= \frac{1}{42} (2 x + 1)^{21}

Add a constant to the solution

= \frac{1}{42} (2 x + 1)^{21} + C

\int (x e^{- x})^{2} d x

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What is the integral of (2x+1)^{20} ?
The integral of (2x+1)^{20} is 1/42 (2x+1)^{21}+C