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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int \frac{u ^{3}}{2} d u

= 2 \cdot \int \frac{u ^{3}}{2} d u

Take the constant out:

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

= 2 \cdot \frac{1}{2} \cdot \int u^{3} d u

Apply the Power Rule

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

= 2 \cdot \frac{1}{2} \cdot \frac{u ^{4}}{4}

Substitute back

u = x^{2} - 3

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

Simplify

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

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

Add a constant to the solution

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

\int x^{k} d x

(x + 2) (x^{3} - x)

\frac{d}{d x} (\frac{x ^{3}}{x})

\int \frac{1}{( 16 - x ^{2} ) ^{3}} d x

\frac{x ^{2} + 4 x + 3}{x}

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