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

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

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integral of 5(5x-1)^3

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

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

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

= 5 \cdot \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

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

Apply the Power Rule

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

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

Substitute back

u = 5 x - 1

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

Simplify

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

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

Add a constant to the solution

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

\int e^{x} + x cos (y) d y

n = 1 \sum \infty n^{3} \cdot e^{- n^{2}}

x \to 4 lim (5 x^{2} - 2 x)

\frac{4 + t}{t e ^{t}}

\int \frac{1}{t ^{3} t ^{2} - 1} d t