What is the integral of 12(3x+1)^3 ?

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

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integral of 12(3x+1)^3

\int 12 (3 x + 1)^{3} d x

(3 x + 1)^{4} + C

Solution steps

\int 12 (3 x + 1)^{3} d x

Take the constant out:

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

= 12 \cdot \int (3 x + 1)^{3} d x

Apply u-substitution

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

= 12 \cdot \int \frac{u ^{3}}{3} d u

Take the constant out:

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

= 12 \cdot \frac{1}{3} \cdot \int u^{3} d u

Apply the Power Rule

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

= 12 \cdot \frac{1}{3} \cdot \frac{u ^{4}}{4}

Substitute back

u = 3 x + 1

= 12 \cdot \frac{1}{3} \cdot \frac{( 3 x + 1 ) ^{4}}{4}

Simplify

12 \cdot \frac{1}{3} \cdot \frac{( 3 x + 1 ) ^{4}}{4} : (3 x + 1)^{4}

= (3 x + 1)^{4}

Add a constant to the solution

= (3 x + 1)^{4} + C

\int (\frac{3}{2}) d x

\frac{d}{d x} ((x^{2} + 1)^{3} (x^{2} + 2)^{6})

\int_{2}^{3} \frac{14}{3 - x} d x

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

x \to 0 lim (\frac{2 x + 1}{x ^{3}})