What is the integral of 12x(3x^2-1)^3 ?

The integral of 12x(3x^2-1)^3 is 1/2 (3x^2-1)^4+C

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integral of 12x(3x^2-1)^3

\int 12 x (3 x^{2} - 1)^{3} d x

\frac{1}{2} (3 x^{2} - 1)^{4} + C

Solution steps

\int 12 x (3 x^{2} - 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 x (3 x^{2} - 1)^{3} d x

Apply u-substitution

: \int \frac{u ^{3}}{6} d u

= 12 \cdot \int \frac{u ^{3}}{6} d u

Take the constant out:

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

= 12 \cdot \frac{1}{6} \cdot \int u^{3} d u

Apply the Power Rule

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

= 12 \cdot \frac{1}{6} \cdot \frac{u ^{4}}{4}

Substitute back

u = 3 x^{2} - 1

= 12 \cdot \frac{1}{6} \cdot \frac{( 3 x ^{2} - 1 ) ^{4}}{4}

Simplify

12 \cdot \frac{1}{6} \cdot \frac{( 3 x ^{2} - 1 ) ^{4}}{4} : \frac{1}{2} (3 x^{2} - 1)^{4}

= \frac{1}{2} (3 x^{2} - 1)^{4}

Add a constant to the solution

= \frac{1}{2} (3 x^{2} - 1)^{4} + C

f (x) = 3 - 6 x

\frac{d}{d x} (c e^{- 2 x})

\int_{1}^{2} x^{2} - 1 d x

y^{^{'}} = \frac{( 1 - 2 x )}{y}

(x + 5) e^{2 x} + 7

What is the integral of 12x(3x^2-1)^3 ?
The integral of 12x(3x^2-1)^3 is 1/2 (3x^2-1)^4+C