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

\int_{1}^{6} (2 x^{3} + 5 x^{2} + 2 x + 8) d x

x \to 3 + lim (\frac{1}{6 - 2 x})

f (x) = 1.1 x + e^{x}, at x = 0

\int \frac{x ^{2}}{x ^{4} - 6 x ^{2} - 27} d x

y y^{^{'}} + 36 x = 0