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

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

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

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

\frac{x ^{12}}{4} + x^{9} + \frac{3 x ^{6}}{2} + x^{3} + C

Solution steps

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

Take the constant out:

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

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

Expand

x^{2} (x^{3} + 1)^{3} : x^{11} + 3 x^{8} + 3 x^{5} + x^{2}

= 3 \cdot \int x^{11} + 3 x^{8} + 3 x^{5} + x^{2} 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

= 3 (\int x^{11} d x + \int 3 x^{8} d x + \int 3 x^{5} d x + \int x^{2} d x)

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

\int 3 x^{8} d x = \frac{x ^{9}}{3}

\int 3 x^{5} d x = \frac{x ^{6}}{2}

\int x^{2} d x = \frac{x ^{3}}{3}

= 3 (\frac{x ^{12}}{12} + \frac{x ^{9}}{3} + \frac{x ^{6}}{2} + \frac{x ^{3}}{3})

Simplify

3 (\frac{x ^{12}}{12} + \frac{x ^{9}}{3} + \frac{x ^{6}}{2} + \frac{x ^{3}}{3}) : \frac{x ^{12}}{4} + x^{9} + \frac{3 x ^{6}}{2} + x^{3}

= \frac{x ^{12}}{4} + x^{9} + \frac{3 x ^{6}}{2} + x^{3}

Add a constant to the solution

= \frac{x ^{12}}{4} + x^{9} + \frac{3 x ^{6}}{2} + x^{3} + C

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