What is the integral of (3x+2)^4 ?

The integral of (3x+2)^4 is 1/15 (3x+2)^5+C

integral of (3x+2)^4

\int (3 x + 2)^{4} d x

\frac{1}{15} (3 x + 2)^{5} + C

Solution steps

\int (3 x + 2)^{4} d x

Apply u-substitution

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

Substitute back

u = 3 x + 2

= \frac{1}{3} \cdot \frac{( 3 x + 2 ) ^{5}}{5}

Simplify

\frac{1}{3} \cdot \frac{( 3 x + 2 ) ^{5}}{5} : \frac{1}{15} (3 x + 2)^{5}

= \frac{1}{15} (3 x + 2)^{5}

Add a constant to the solution

= \frac{1}{15} (3 x + 2)^{5} + C

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

\frac{d}{d x} (\frac{4 x - 12}{( x - 2 ) ^{2}})

x \to 3 - lim (3 - x)

\int x^{7} x^{4} + 3 d x

t an g e n t f (x) = \frac{3}{x + 1}, (2, 1)