What is the integral of (4x+1)^3 ?

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

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integral of (4x+1)^3

\int (4 x + 1)^{3} d x

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

Solution steps

\int (4 x + 1)^{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 + 1

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

Simplify

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

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

Add a constant to the solution

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

- sec (\frac{1}{x} + 7 x)

\frac{d}{d x} (x e^{- 2 x^{2}})

\frac{\partial}{\partial x} (7 e^{x y})

\frac{d}{d x} (\frac{x}{ln ( x )})