What is the integral of (5x-1)^3 ?

The integral of (5x-1)^3 is 1/20 (5x-1)^4+C

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integral of (5x-1)^3

\int (5 x - 1)^{3} d x

\frac{1}{20} (5 x - 1)^{4} + C

Solution steps

\int (5 x - 1)^{3} d x

Apply u-substitution

: \int \frac{u ^{3}}{5} d u

= \int \frac{u ^{3}}{5} d u

Take the constant out:

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

= \frac{1}{5} \cdot \int u^{3} d u

Apply the Power Rule

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

= \frac{1}{5} \cdot \frac{u ^{4}}{4}

Substitute back

u = 5 x - 1

= \frac{1}{5} \cdot \frac{( 5 x - 1 ) ^{4}}{4}

Simplify

\frac{1}{5} \cdot \frac{( 5 x - 1 ) ^{4}}{4} : \frac{1}{20} (5 x - 1)^{4}

= \frac{1}{20} (5 x - 1)^{4}

Add a constant to the solution

= \frac{1}{20} (5 x - 1)^{4} + C

\frac{3 x ^{2}}{( 3 - 2 x ) ^{4}}

y^{^{''}} + 16 y = 2 sin (2 t), y (0) = 1, y^{^{'}} (0) = 0

\int (2 t e^{t}) d t

y^{4} - 4 y^{2} = x^{4} - 9 x^{2}, (- 3, 2)

y = e^{4 x} cos (7 x)