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

The integral of (3x-1)^4 is 1/15 (3x-1)^5+C

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integral of (3x-1)^4

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

\frac{1}{15} (3 x - 1)^{5} + C

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Apply the Power Rule

: \frac{u ^{5}}{5}

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

Substitute back

u = 3 x - 1

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

Simplify

\frac{1}{3} \cdot \frac{( 3 x - 1 ) ^{5}}{5} : \frac{1}{15} (3 x - 1)^{5}

= \frac{1}{15} (3 x - 1)^{5}

Add a constant to the solution

= \frac{1}{15} (3 x - 1)^{5} + C

\int_{0}^{1} \frac{37}{x ^{p}} d x

\frac{d}{d x} (e^{3 x^{2}} \cdot 6 x)

y = 2 x^{3} + 3 x^{2} - 12 x + 6

\frac{d ^{2} y}{d t ^{2}} + 3 \frac{d y}{d t} + y = 1

x \to 2 lim (- 10 x + (10))