What is the integral of 6(x^2+1)(x^3+3x)^3 ?

The integral of 6(x^2+1)(x^3+3x)^3 is 6((x^{12}}{12}+x^{10}+\frac{9x^8)/2+9x^6+(27x^4)/4)+C

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integral of 6(x^2+1)(x^3+3x)^3

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

6 (\frac{x ^{12}}{12} + x^{10} + \frac{9 x ^{8}}{2} + 9 x^{6} + \frac{27 x ^{4}}{4}) + C

Solution steps

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

Take the constant out:

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

= 6 \cdot \int (x^{2} + 1) (x^{3} + 3 x)^{3} d x

Expand

(x^{2} + 1) (x^{3} + 3 x)^{3} : x^{11} + 10 x^{9} + 36 x^{7} + 54 x^{5} + 27 x^{3}

= 6 \cdot \int x^{11} + 10 x^{9} + 36 x^{7} + 54 x^{5} + 27 x^{3} 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^{11} d x + \int 10 x^{9} d x + \int 36 x^{7} d x + \int 54 x^{5} d x + \int 27 x^{3} d x)

\int x^{11} d x = \frac{x ^{12}}{12}

\int 10 x^{9} d x = x^{10}

\int 36 x^{7} d x = \frac{9 x ^{8}}{2}

\int 54 x^{5} d x = 9 x^{6}

\int 27 x^{3} d x = \frac{27 x ^{4}}{4}

= 6 (\frac{x ^{12}}{12} + x^{10} + \frac{9 x ^{8}}{2} + 9 x^{6} + \frac{27 x ^{4}}{4})

Add a constant to the solution

= 6 (\frac{x ^{12}}{12} + x^{10} + \frac{9 x ^{8}}{2} + 9 x^{6} + \frac{27 x ^{4}}{4}) + C

What is the integral of 6(x^2+1)(x^3+3x)^3 ?
The integral of 6(x^2+1)(x^3+3x)^3 is 6((x^{12}}{12}+x^{10}+\frac{9x^8)/2+9x^6+(27x^4)/4)+C