What is the integral of (2x+1)^{99} ?

The integral of (2x+1)^{99} is 1/200 (2x+1)^{100}+C

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integral of (2x+1)^{99}

\int (2 x + 1)^{99} d x

\frac{1}{200} (2 x + 1)^{100} + C

Solution steps

\int (2 x + 1)^{99} d x

Apply u-substitution

= \int \frac{u ^{99}}{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^{99} d u

Apply the Power Rule

= \frac{1}{2} \cdot \frac{u ^{100}}{100}

Substitute back

u = 2 x + 1

= \frac{1}{2} \cdot \frac{( 2 x + 1 ) ^{100}}{100}

Simplify

\frac{1}{2} \cdot \frac{( 2 x + 1 ) ^{100}}{100} : \frac{1}{200} (2 x + 1)^{100}

= \frac{1}{200} (2 x + 1)^{100}

Add a constant to the solution

= \frac{1}{200} (2 x + 1)^{100} + C

What is the integral of (2x+1)^{99} ?
The integral of (2x+1)^{99} is 1/200 (2x+1)^{100}+C