What is the integral of (x^3+6x)(6x^2+12) ?

The integral of (x^3+6x)(6x^2+12) is x^6+12x^4+36x^2+C

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integral of (x^3+6x)(6x^2+12)

\int (x^{3} + 6 x) (6 x^{2} + 12) d x

x^{6} + 12 x^{4} + 36 x^{2} + C

Solution steps

\int (x^{3} + 6 x) (6 x^{2} + 12) d x

(x^{3} + 6 x) (6 x^{2} + 12) = 6 (x^{3} + 6 x) (x^{2} + 2)

= \int 6 (x^{3} + 6 x) (x^{2} + 2) 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} + 6 x) (x^{2} + 2) d x

Expand

(x^{3} + 6 x) (x^{2} + 2) : x^{5} + 8 x^{3} + 12 x

= 6 \cdot \int x^{5} + 8 x^{3} + 12 x d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 6 (\int x^{5} d x + \int 8 x^{3} d x + \int 12 x d x)

\int x^{5} d x = \frac{x ^{6}}{6}

\int 8 x^{3} d x = 2 x^{4}

\int 12 x d x = 6 x^{2}

= 6 (\frac{x ^{6}}{6} + 2 x^{4} + 6 x^{2})

Simplify

6 (\frac{x ^{6}}{6} + 2 x^{4} + 6 x^{2}) : x^{6} + 12 x^{4} + 36 x^{2}

= x^{6} + 12 x^{4} + 36 x^{2}

Add a constant to the solution

= x^{6} + 12 x^{4} + 36 x^{2} + C

What is the integral of (x^3+6x)(6x^2+12) ?
The integral of (x^3+6x)(6x^2+12) is x^6+12x^4+36x^2+C