What is the integral of (2x+1)^8 ?

The integral of (2x+1)^8 is 1/18 (2x+1)^9+C

Popular >

integral of (2x+1)^8

\int (2 x + 1)^{8} d x

\frac{1}{18} (2 x + 1)^{9} + C

Solution steps

\int (2 x + 1)^{8} d x

Apply u-substitution

: \int \frac{u ^{8}}{2} d u

= \int \frac{u ^{8}}{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^{8} d u

Apply the Power Rule

: \frac{u ^{9}}{9}

= \frac{1}{2} \cdot \frac{u ^{9}}{9}

Substitute back

u = 2 x + 1

= \frac{1}{2} \cdot \frac{( 2 x + 1 ) ^{9}}{9}

Simplify

\frac{1}{2} \cdot \frac{( 2 x + 1 ) ^{9}}{9} : \frac{1}{18} (2 x + 1)^{9}

= \frac{1}{18} (2 x + 1)^{9}

Add a constant to the solution

= \frac{1}{18} (2 x + 1)^{9} + C

\int \frac{( 1 + x )}{( 1 - x )} d x

\int (2 - x) d x

\int \frac{6}{( 3 + 6 x ) ^{2}} d x

\int \frac{- e ^{x}}{e ^{x} + 1} d x

\int (x^{2} - 6)^{5} (2 x) d x