What is the integral of (2z+1)^3 ?

The integral of (2z+1)^3 is 1/8 (2z+1)^4+C

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integral of (2z+1)^3

\int (2 z + 1)^{3} d z

\frac{1}{8} (2 z + 1)^{4} + C

Solution steps

\int (2 z + 1)^{3} d z

Apply u-substitution

: \int \frac{u ^{3}}{2} d u

= \int \frac{u ^{3}}{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^{3} d u

Apply the Power Rule

: \frac{u ^{4}}{4}

= \frac{1}{2} \cdot \frac{u ^{4}}{4}

Substitute back

u = 2 z + 1

= \frac{1}{2} \cdot \frac{( 2 z + 1 ) ^{4}}{4}

Simplify

\frac{1}{2} \cdot \frac{( 2 z + 1 ) ^{4}}{4} : \frac{1}{8} (2 z + 1)^{4}

= \frac{1}{8} (2 z + 1)^{4}

Add a constant to the solution

= \frac{1}{8} (2 z + 1)^{4} + C

y = (1 + 3 x)^{10}, (0, 1)

x \to 0 lim (2 e^{5 x})

\int_{0}^{7} - 30 x^{2} + 340 x d x

x \to + 0 lim (\frac{1}{x - 1})

(x^{2} - y^{2}) d x + 3 x y d y = 0