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

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

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integral of (2x+3)^3

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

\frac{1}{8} (2 x + 3)^{4} + C

Solution steps

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

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 x + 3

= \frac{1}{2} \cdot \frac{( 2 x + 3 ) ^{4}}{4}

Simplify

\frac{1}{2} \cdot \frac{( 2 x + 3 ) ^{4}}{4} : \frac{1}{8} (2 x + 3)^{4}

= \frac{1}{8} (2 x + 3)^{4}

Add a constant to the solution

= \frac{1}{8} (2 x + 3)^{4} + C

\frac{d}{d x} (\frac{25 x ^{3}}{3} + 10 x^{2} + 4 x - 7)

f^{^{'}} (x) = e^{- 2 x}

\frac{d}{d x} (\frac{2 x + sin ( x )}{2 x + cos ( x )})

(1, 10), (24, - 26)

x \to 4 - lim (5 - x)