What is the integral of (4x+3)^3 ?

The integral of (4x+3)^3 is 1/16 (4x+3)^4+C

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integral of (4x+3)^3

\int (4 x + 3)^{3} d x

\frac{1}{16} (4 x + 3)^{4} + C

Solution steps

\int (4 x + 3)^{3} d x

Apply u-substitution

: \int \frac{u ^{3}}{4} d u

= \int \frac{u ^{3}}{4} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{4} \cdot \int u^{3} d u

Apply the Power Rule

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

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

Substitute back

u = 4 x + 3

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

Simplify

\frac{1}{4} \cdot \frac{( 4 x + 3 ) ^{4}}{4} : \frac{1}{16} (4 x + 3)^{4}

= \frac{1}{16} (4 x + 3)^{4}

Add a constant to the solution

= \frac{1}{16} (4 x + 3)^{4} + C

f (x) = e^{x^{2}} \cdot 2 x

\int 5 x^{3} 1 - x^{2} d x

\frac{\partial}{\partial x} (2 x^{2} - 4 x y + x yz)

\frac{\partial}{\partial x} (\frac{( x ( y + 3 ) )}{x ^{2} + y})

x^{4} e^{y} + (x + 1) y \frac{d y}{d x} = 0