What is the integral of 3(3x+12)^4 ?

The integral of 3(3x+12)^4 is 1/5 (3x+12)^5+C

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integral of 3(3x+12)^4

\int 3 (3 x + 12)^{4} d x

\frac{1}{5} (3 x + 12)^{5} + C

Solution steps

\int 3 (3 x + 12)^{4} d x

Take the constant out:

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

= 3 \cdot \int (3 x + 12)^{4} d x

Apply u-substitution

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

= 3 \cdot \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

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

Apply the Power Rule

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

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

Substitute back

u = 3 x + 12

= 3 \cdot \frac{1}{3} \cdot \frac{( 3 x + 12 ) ^{5}}{5}

Simplify

3 \cdot \frac{1}{3} \cdot \frac{( 3 x + 12 ) ^{5}}{5} : \frac{1}{5} (3 x + 12)^{5}

= \frac{1}{5} (3 x + 12)^{5}

Add a constant to the solution

= \frac{1}{5} (3 x + 12)^{5} + C

y^{^{'}} = 5 y

\frac{d}{d x} (arctan (\frac{x}{1 - x}))

z \to 0 lim ((2 z - 8)^{\frac{1}{3}})

\frac{d}{d x} (\frac{2}{3} (1 - x)^{\frac{3}{2}})

y^{^{'}} = - 2 x tan (y), y (0) = \frac{π}{2}

What is the integral of 3(3x+12)^4 ?
The integral of 3(3x+12)^4 is 1/5 (3x+12)^5+C