What is the integral of 6x(3x^2-2)^4 ?

The integral of 6x(3x^2-2)^4 is 1/5 (3x^2-2)^5+C

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integral of 6x(3x^2-2)^4

\int 6 x (3 x^{2} - 2)^{4} d x

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

Solution steps

\int 6 x (3 x^{2} - 2)^{4} d x

Take the constant out:

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

= 6 \cdot \int x (3 x^{2} - 2)^{4} d x

Apply u-substitution

: \int \frac{u ^{4}}{6} d u

= 6 \cdot \int \frac{u ^{4}}{6} d u

Take the constant out:

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

= 6 \cdot \frac{1}{6} \cdot \int u^{4} d u

Apply the Power Rule

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

= 6 \cdot \frac{1}{6} \cdot \frac{u ^{5}}{5}

Substitute back

u = 3 x^{2} - 2

= 6 \cdot \frac{1}{6} \cdot \frac{( 3 x ^{2} - 2 ) ^{5}}{5}

Simplify

6 \cdot \frac{1}{6} \cdot \frac{( 3 x ^{2} - 2 ) ^{5}}{5} : \frac{1}{5} (3 x^{2} - 2)^{5}

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

Add a constant to the solution

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

\int_{0}^{\frac{1}{5}} 4 arcsin (5 y) d y

y = 3 x - 4 x^{2}, (- 2, - 22)

\frac{\partial}{\partial x} (e^{- x y} cos (y))

\frac{1}{s ^{2} ( s ^{2} + 3 s + 10 )}

\frac{d}{d x} (\frac{4 e ^{2 x} - 2 e ^{x} + 1}{2 e ^{2 x}})

What is the integral of 6x(3x^2-2)^4 ?
The integral of 6x(3x^2-2)^4 is 1/5 (3x^2-2)^5+C